On Sun, Jan 19, 2025 at 12:21 PM Alan Grayson <agrayson2...@gmail.com>
wrote:
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> On Saturday, January 18, 2025 at 9:30:47 PM UTC-7 Jesse Mazer wrote:
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> On Sat, Jan 18, 2025 at 8:25 PM Alan Grayson <agrays...@gmail.com> wrote:
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> On Saturday, January 18, 2025 at 1:25:08 PM UTC-7 Jesse Mazer wrote:
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> On Sat, Jan 18, 2025 at 2:55 PM Alan Grayson <agrays...@gmail.com> wrote:
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> On Saturday, January 18, 2025 at 11:50:11 AM UTC-7 Jesse Mazer wrote:
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> On Sat, Jan 18, 2025 at 1:19 PM Alan Grayson <agrays...@gmail.com> wrote:
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> On Saturday, January 18, 2025 at 10:24:01 AM UTC-7 Jesse Mazer wrote:
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> On Sat, Jan 18, 2025 at 12:09 PM Alan Grayson <agrays...@gmail.com> wrote:
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> On Saturday, January 18, 2025 at 9:15:17 AM UTC-7 Jesse Mazer wrote:
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> On Sat, Jan 18, 2025 at 9:36 AM Alan Grayson <agrays...@gmail.com> wrote:
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> On Saturday, January 18, 2025 at 7:19:41 AM UTC-7 Jesse Mazer wrote:
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> On Sat, Jan 18, 2025 at 9:09 AM Alan Grayson <agrays...@gmail.com> wrote:
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> On Thursday, January 16, 2025 at 5:52:52 PM UTC-7 Jesse Mazer wrote:
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> On Thu, Jan 16, 2025 at 7:33 PM Alan Grayson <agrays...@gmail.com> wrote:
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> On Thursday, January 16, 2025 at 2:39:55 PM UTC-7 Jesse Mazer wrote:
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> On Thu, Jan 16, 2025 at 2:43 PM Alan Grayson <agrays...@gmail.com> wrote:
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> On Thursday, January 16, 2025 at 11:36:48 AM UTC-7 Jesse Mazer wrote:
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> On Tue, Jan 14, 2025 at 12:02 AM Alan Grayson <agrays...@gmail.com> wrote:
>
> Using the LT, we have the following transformations of Length, Time, and
> Mass, that is,
> x --->x', t ---> t', m ---> m'
>
>
> The length contraction equation is not part of the Lorentz transformation
> equations, the x --> x' equation in the LT is just about the position
> coordinate assigned to a *single* event in each frame. The length
> contraction equation can be derived from the LT but only by considering
> worldlines of the front and back of an object, and looking at *pairs* of
> events (one on each of the two worldlines) which are simultaneous in each
> frame--length in a given frame is just defined as the difference in
> position coordinate between the front and back of an object at a single
> time-coordinate in that frame, so it requires looking at a pair of events
> that are simultaneous in that frame. The result is that for any inertial
> object, it has its maximum length L in the frame where the object is at
> rest (the object's own 'rest frame'), and a shorter length L*sqrt(1 -
> v^2/c^2) in a different frame where the object has nonzero velocity v.
>
> The t ---> t' equation is likewise not the same as the time dilation
> equation, it's just about the time coordinate assigned to a single event in
> each frame, although it has a simpler relation to time dilation since you
> can consider an event on the worldline that passes through the origin where
> both t and t' are equal to 0, and then the time coordinates t and t'
> assigned to some other event E on this worldline tell you the time elapsed
> in each frame between the origin and E. And the LT don't include any mass
> transformation equation.
>
> Jesse
>
>
> You're right of course. TY. I see the LT as giving appearances because,
> say for length contraction, the reduced length is not measured in the
> primed frame, but that is the length measurement from the pov of the
> unprimed or stationary frame.
>
>
> In relativity one does not normally designate any particular frame to be
> the "stationary frame", since all concepts of motion and rest are defined
> in purely relative way; if one has two objects A and B in relative motion,
> one could talk about the frame where A is stationary (A's 'rest frame') or
> the frame where B is stationary (B's rest frame), but that's all. I'm not
> sure what you mean by "the reduced length is not measured in the primed
> frame"--which object's length are you talking about? If A's rest frame is
> the unprimed frame and B's rest frame is the primed frame, then the length
> of object A in the primed frame is reduced relative to its length in its
> own rest frame, i.e. the unprimed frame.
>
>
> *Let's consider a concrete example of a traveler moving at near light
> speed to Andromeda. From the traveler's frame, the distance to Andromeda is
> hugely reduced from its length of 2.5 MLY from the pov of a non-traveling
> observer. This seems to imply that the reduced length is only measured from
> the pov of the traveler, but not from the pov of the non-traveler, because
> of which I describe the measurement from the pov of the traveler as
> APPARENT. Do you agree that the traveler's measurement is apparent because
> the non-traveler measures the distance to Andromeda as unchanged? TY, AG *
>
>
> I don't know what you mean by "apparent", but there is no asymmetry in the
> way Lorentz contraction works in each frame--
>
>
> *I mean, if one uses the LT, to transform from one frame to another frame,
> if the resultant parameters in the latter frame are not actually measured
> in the latter frame, I refer to those measurements as "apparent". What I'm
> stuggling with is what the LT actually results in. Does it tell us what is
> actually measured in the latter frame, or not? AG*
>
>
> Yes, it always tells you what is actually measured in the frame
> that you're transforming into, using that frame's own rulers and clocks.
> You haven't made it at all clear why you suspect otherwise.
>
> Jesse
>
>
> *I suspect otherwise because in the Andromeda problem, using the LT from
> the pov of the rest frame, at rest relative to the Earth, we get length
> contraction in the transformed frame (modeled as a rod moving toward the
> Earth)*
>
>
> I thought the problem was supposed to involve the assumption of a traveler
> going from Earth to Andromeda (who I imagined as riding in a rocket), with
> a rod that is at rest relative to Earth and Andromeda and whose length
> defines the distance between them in each frame? But here you seem to be
> talking about a rod moving relative to the Earth? So now I'm not even clear
> about what physical scenario you are imagining--please lay it out clearly
> by specifying all the physical objects you are imagining in the problem
> (like rocket, rod, Earth, Andromeda), which of them have nonzero velocity
> in the Earth's frame, and the rest length for any object you want to do
> length calculations for.
>
> Jesse
>
>
> *If you prefer, we can place the traveler to Andromeda in a spaceship
> moving with some velocity wrt the Earth, and now imagine a rod whose length
> is the distance from Earth to Andromeda.*
>
>
> So the rod is at rest relative to Earth and Andromeda, as I originally
> understood you to be saying?
>
>
>
> * Since motion is relative, we can imagine the spaceship is at rest, and
> the rod moving toward Earth (since the spaceship is ..imagined as moving
> toward Andromeda).*
>
>
> By "moving toward the Earth" do you mean moving *relative to the Earth*
> (i.e. distance between the Earth and either end of the rod is changing over
> time), or do you just mean that in the spaceship's frame, the rod is always
> moving in the direction of the Earth, but the Earth is moving at the same
> speed in the same direction so the end of the rod always coincides with the
> Earth (or remains at a fixed negligible distance from it)? If the latter,
> "the rod moving toward Earth" seems like confusing terminology for this,
> there would be much less room for confusion if you stuck to the language of
> talking about movement "relative to" a given object or observer as I
> suggested.
>
>
>
> * In the rest frame of the spaceship, the rod is contracted since it
> appears to be **moving. We can use the LT to calculate how much its
> length contracts due to the velocity of the rod. However, as I pointed out
> earlier, this measurement is NOT possible IN the frame of the moving rod**,
> since nothing is moving within this frame.*
>
>
> How much its length contracts in the frame of the spaceship, or in the
> frame of the rod (assuming the rod is at rest relative to Earth/Andromeda)?
>
>
>
> * This is an example of the LT NOT yielding a measurement in the target
> frame (in this case the frame containing the rod), which is an exception to
> the claim that the LT always predicts what a target frame will actually
> measure.*
>
>
> I don't know why you think that, but you'll have to clearly specify which
> frame you want to transform from, and which frame you want to use the LT to
> transform into (the 'target frame'). If you mean starting from the
> coordinates of the rod in the spaceship's rest frame, and using the LT to
> transform into the rest frame of the rod (so the rod's frame is the target
> frame), then if you actually do the algebra of the LT you will get the
> correct result that the rod is *longer* in its own rest frame than it was
> in the spaceship frame, not shorter. And if on the other hand you start
> with the coordinates in the rod's rest frame and use the LT to transform
> into the spaceship frame (so the spaceship frame is the target frame), you
> will get the correct result that the rod is shorter in the spaceship frame
> than in its own rest frame.
>
> Jesse
>
>
> *Let me try again. The spaceship is moving toward Andromeda from the
> Earth. Since motion is relative I can assume the spaceship is at rest, and
> a rod representing the distance from Earth to Andromeda is moving toward
> the Earth, since it is assumed the spaceship was originally moving in the
> opposite direction, toward Andromeda. *
>
>
> You didn't address my question of whether "moving toward the Earth" means
> moving *relative* to the Earth (i.e. rod is moving in Earth's rest frame,
> and other frames like the spaceship frame see the distance between the end
> of the rod and the Earth as changing over time), or if it just means that
> in the rod is moving in the direction of the Earth at the same speed that
> the Earth itself is moving in this frame, so the distance between any point
> on the rod and the Earth is unchanging. If you mean the latter, this is
> confusing terminology, definitely not the sort of thing you would find in
> any relativity textbook. Either way, can you *please* stick to defining
> movement and rest "relative to" specific objects or observers to prevent
> this kind of verbal ambiguity?
>
>
> *I assume, and you can as well, that the Earth is at rest, and the
> spaceship is moving toward Andromeda. Can you assume the Earth is at rest
> in this model and not allow us to get into a discussion of what "at rest"
> means? AG*
>
>
> I think this way of speaking just leads to confusion, which is why I'd
> prefer that you not designate any particular frame as being "at rest" and
> talk exclusively in a relative way about "the rest frame of the Earth" and
> "the rest frame of the spaceship". Even if you are stubborn about this and
> really want to designate one frame as being "at rest" and one as "moving",
> you are not even being consistent in your designation, since in your post
> immediately before this you said "Since motion is relative I can assume the
> spaceship is at rest". There would be much less room for confusion if you
> would just do as I suggest and use "rest" and "moving" in a relative way
> ('the rest frame of X', 'X is moving relative to Y' etc.), would you be
> willing to do this for my sake or do you absolutely refuse? Please answer
> this question directly, I've asked before and you've simply not responded.
>
> Also, note that my question above was specifically about the *rod", which
> you didn't mention at all in your response above--I was asking about the
> meaning of your statement that the rod is "moving toward the Earth", can
> you please just answer yes or no if the rod is meant to be at rest relative
> to the Earth, and therefore moving at the same velocity as the Earth in the
> spaceship rest frame?
>
>
> *I was thinking the spaceship is relatively at rest,*
>
>
> "Relatively at rest" is a meaningless phrase unless you specify what
> *specific object* it is at rest relative to. Are you so attached to your
> own non-standard way of talking that you refuse to just use the
> formulations "at rest relative to X" and "moving relative to X" (where X is
> some object in the problem) as a small concession towards being understood
> by others? I have asked if you are willing to do this a bunch of times,
> last time I even asked you "please answer this question directly", but you
> continue to just ignore it. I'm not going to continue the discussion with
> you if you refuse to do me this simple courtesy, without it I genuinely
> can't understand what scenario you're envisioning.
>
>
Here you have again ignored the general question I ask about whether you
are willing to phrase all your comments about rest/motion in a way that
makes explicit what frame the statement is relative to (like 'moving
relative to the Earth' which makes clear we are talking about motion in the
Earth's rest frame). Are you willing to do this, yes or no? If you won't
give me an answer, I'm not going to continue this discussion with you.
>
>
> * the rod is moving toward the Earth,*
>
>
> And I likewise asked several times to specify if you mean the rod is
> moving *relative* to the Earth (i.e. the distance between the rod and the
> Earth is changing over time), or if you just mean it's moving in the
> direction of the Earth but at a fixed distance (i.e. it's at rest relative
> to the Earth). My question last time was "can you please just answer yes or
> no if the rod is meant to be at rest relative to the Earth, and therefore
> moving at the same velocity as the Earth in the spaceship rest frame?" If
> you aren't willing to answer simple yes-or-no questions like this, then
> again I'm not going to continue the discussion.
>
>
> * and the LT is used by the spaceship to calculate the contraction of the
> rod. *
>
>
> Contraction of the rod in which object's rest frame?
>
>
>
> *I tried to make it clear, very clear, but obviously I failed. Let's try
> this; since there's general agreement in the physics community that
> traveling very fast to Andromeda causes its distance from Earth to
> contract. So you tell me; from which frame would you'd like to measure the
> contraction?*
>
>
> In the spaceship frame, obviously.
>
>
>
> *And since, as Brent just remarked, in the contracted frame, nothing that
> the LT determined, is measurable in the contracted frame, my claim is
> proven;*
>
>
> "Contracted frame" is another unclear phrase, the distance from Earth to
> Andromeda is contracted in the spaceship frame but the spaceship itself is
> contracted in the Earth/Andromeda frame. *If* by "contracted frame" you
> mean the spaceship frame where the distance from Earth to Andromeda is
> contracted relative to the distance between them in their rest frame, then
> you need to understand that what is measured in the spaceship frame (using
> a system of rulers and clocks at rest in that frame, the clocks
> synchronized using the Einstein convention) is exactly what would be
> predicted if you started with the coordinates in some other frame and use
> the LT to transform into the spaceship frame. If you think Brent is saying
> otherwise, I can guarantee you've just misunderstood him.
>
>
> *The spaceship is moving toward Andromeda at some velocity v wrt the
> Earth. The rod is moving with velocity -v toward the Earth,*
>
While "velocity v wrt the Earth" is the sort of phrase I asked for, I have
already commented in three previous posts about the ambiguity of your
statement that the rod is moving "toward the Earth" and asked you to
clarify, which you still have not done, instead you're just repeating the
same ambiguous phrase. As I said before, "moving toward the Earth" could
either mean "moving relative to the Earth" or it could mean "moving in the
direction of the Earth's position, as seen in the spaceship's frame where
both rod and Earth have the same velocity -v" (in the latter case where
they both have the same velocity in the spaceship frame, this would imply
the rod is at rest in the Earth's own frame). Will you please answer my
question about whether the rod is moving *relative to* (or wrt) the Earth,
yes or no?
In general it's frustrating that when I ask you several pointed questions,
especially ones where I ask for a simple yes-or-no answer, you just try to
restate your overall scenario in a way that you perhaps vaguely think
addresses everything, without actually quoting my questions and giving
individual responses to them. Can you please answer each question
individually? I note that you also completely ignored the final question in
my last post about the symmetry of rest vs. moving between frames (see my
last comment below).
[BTW, an alternative way you could avoid all this verbal ambiguity would be
to give an actual numerical example where you state the position as a
function of time for each object in the scenario. For example, say the
spaceship and Earth/Andromeda have a relative speed of 0.6c so that the
distance from Earth to Andromeda is contracted by a factor of 0.8 in the
spaceship rest frame, meaning if the distance is 2.5 Gly in the
Earth/Andromeda rest frame (Gly = Giga-light-years, so 2.5 Gly = 2.5
million light years), then the distance between them would be 2 Gly in the
ship rest frame. In that case, if we also use units of Giga-years for time
so that c=1, then you could say something like "in the ship's rest frame,
at t=0 the initial conditions are that the ship is at position x = 0, the
Earth is also at x = 0, and Andromeda is at position x = -2, and the ship's
position as a function of time in this frame is x(t) = 0, the Earth's
position as a function of time is x(t) = 0.6*t, and the Andromeda galaxy's
position as a function of time is x(t) = 0.6*t - 2". Then if you also gave
the corresponding equations of motion for the front and back of the rod in
the ship's rest frame (I'm still not clear on whether they'd be identical
to the equations of motion for Earth and Andromeda, so that one end of the
rod always coincides with Earth and the other end always coincides with
Andromeda, or if they'd be different) then there would be no ambiguity
about what the rod is supposed to be at rest relative to and what it is
moving relative to.]
> * while the spaceship is now at rest wrt the Earth. I am replacing a
> moving spacecraft with a moving rod of known initial length, the distance
> between Andromeda the the Earth, moving in the opposite direction. Then I
> am doing a LT from the stationary frame of the spacecraft to the moving
> frame of the rod, to calculate the rod's contraction from the pov of the
> frame of the spacecraft. I hope I have met your criterion for defining
> these frames of reference clearly.*
>
> *But all this is unnecessary to prove my point; that in the frame of the
> rod, observers CANNOT measure its length contraction (or time dilation).
> This FACT is universally known and not in dispute, yet at the same time
> many people who claim to understand relativity assert that in the target
> frame of the LT, that is, in this case, in the frame of the rod, observers
> CAN measure what the LT implies. ThIs is FALSE! *
>
The LT simply doesn't "imply" there would be any length contraction of the
rod "in the frame of the rod", so your premise here is completely wrong (if
you had an actual numerical example with the equations of motion x(t)
initially stated in the spaceship rest frame, as I suggested above, you
could plug them into the LT equations directly and see what they predict
about the equations of motion x'(t') in the rod's frame--I assume you have
not actually done such an algebraic exercise and are just relying on some
confused verbal argument to get the wrong idea that the LT would predict
contraction of the rod in the rod's own frame). Whatever the LT implies
about lengths/times in a specific inertial frame, it always corresponds
exactly to what would actually be measured using a system of rulers and
clocks which are at rest in that frame (the clocks synchronized by the
Einstein convention), no exceptions.
> *Of course, as you point out, from the pov of the rod frame, the
> spacecraft is contracted and its clock is dilated, but this fact is
> irrelevant to what I am alleging. Finally, I have NOT misunderstood Brent's
> recent comments on this very thread. You can read them yourself to verify
> my claim. Look at two or three of his recent short posts. I don't get it.
> Why do people who understand relativity keep affirming something -- what
> can be MEASURED in the target frame of the LT -- which isn't true? *
>
> *Another thing worth considering about the parking paradox: it's clear, as
> Clark pointed out not so long ago, that the paradox is caused by the
> assumption of universal time, specifically that the car fits and doesn't
> fit, AT THE SAME TIME. Using the disagreement about simultaneity this error
> is corrected and allegedly the paradox goes away. Yet Brent claims his
> plots show that the fitting and not fitting occur a the same time. I asked
> him more than once to explain this, but he hasn't replied. Do you know
> what's going on on this issue? *
>
> *AG*
>
>
>
>
> * that the results of the LT do NOT tell us what the target frame (in this
> case the frame of the contracted length) will measure. QED! AG *
>
>
>
> *Now, from the pov of the rest frame, the spaceship, imagine a LT to
> determine the contraction of the rod.*
>
>
> Contraction of the rod in what frame?
>
>
> *I was explicit; from the rest frame of spaceship, apply the LT to
> determine the contraction of the rod, assuming you know its original
> length. Now ask yourself this question; in the rod frame, does the
> contraction information you've received from the LT correspond to any
> measurement result in the rod frame? (Answer; of course NOT!).*
>
>
> Assuming we start from the coordinates in the spaceship's rest frame and
> transform into the Earth's rest frame (and assuming the rod is at rest
> relative to the Earth), you will get the conclusion that the rod is LONGER
> in the Earth's rest frame than in the spaceship's rest frame, so I don't
> know what you mean by "the contraction information you've received from the
> LT".
>
>
>
> * This is what I've been trying to demonstrate; results given by the LT do
> NOT always give us what can be measured in the target frame of the LT, in
> this case the rod frame. AG (PS; I sent you a message on Facebook. You can
> reply here if you want.)*
>
>
> Are you saying that if we start with the coordinates of the ends of the
> rod in the spaceship frame (I think the rod is supposed to be moving in
> this frame, correct me if I'm wrong), and then use the LT to find the
> coordinates of the ends of the rod in the rod's own rest frame, you think
> we'd get the false prediction that the rod is contracted in the rod's rest
> frame?
>
>
> *The spaceship is in one frame, at relative rest WRT the rod; the rod is
> in another frame, in relative motion WRT to the spaceship,*
>
>
> How can the spaceship be at rest relative to the rod while the rod is in
> motion relative to the spaceship? Rest/motion are always symmetric in
> relativity as in classical mechanics--if X has a speed v in Y's rest frame
> (including v=0 in the case where X is at rest relative to Y), then Y always
> has the same speed v in X's rest frame. Did you just type this out wrong or
> are you really confused on this point?
>
>
> You also did not answer my question above--that point about two objects
> always having symmetrical judgments about whether the other object is at
> rest or moving (and each judging the other's speed to be identical) is a
> very basic and important one, if you don't understand this it's likely to
> lead to endless confusion. So if you do want to continue the discussion and
> are willing to answer yes-or-no questions, then please answer yes or no if
> you understand that if the spaceship is at rest relative to the rod, that
> guarantees that the rod is at rest relative to the spaceship.
>
>
Can you please address the individual yes-or-no question above, not as part
of a longer statement mushing all your comments together but as a distinct
answer to what I ask here?
Jesse
>
>
>
>
> * at the velocity which the spaceship was originally moving toward
> Andromeda. We know the rod's initial length and we calculate its
> contraction from the spaceship's frame, using the rod's relative velocity.
> AG*
>
>
> If so this is simply WRONG, that's not what you would find if you actually
> did the algebra of the Lorentz transformation equations. If you mean
> something different you need to be more clear about what frame we are
> starting with, what is the "target frame" we are transforming into using
> the LT, and which of these frames you want to know the "contraction of the
> rod" in.
>
> Jesse
>
>
>
> * Note that in the frame of the moving rod, its contraction, predicted by
> the LT, cannot be measured because although the rod is moving, nothing
> within the rod frame is moving. This, IMO, contradicts the claim that the
> LT always yields what is actually measured in the target frame of the LT,
> in this case the frame of the rod. OK? AG*
>
>
>
>
> * I think time dilation is an example of the opposite; a measurement which
> is realized in the target frame. AG *
>
>
>
>
>
> *but never length contraction as what's observed IN the rod frame. I
> realize that length contraction requires something moving, so it's not
> reasonable to expect this, although it does show there's an exception to
> what you allege the LT does. AG*
>
>
>
>
>
> if we assume there is a frame A where Milky Way and Andromeda are both at
> rest (ignoring the fact that in reality they have some motion relative to
> one another), and another frame B where the rocket ship of the traveler is
> at rest, then in frame B the Milky Way/Andromeda distance is shortened
> relative to the distance in their rest frame, and the rocket has its
> maximum length; in frame A the the rocket's length is shortened relative to
> its length in its rest frame, and the Milky Way/Andromeda distance has its
> maximum value. The only asymmetry here is in the choice of the two things
> to measure the length of (the distance between the Milky Way and Andromeda
> in their rest frame is obviously huge compared to the rest length of a
> rocket moving between them), the symmetry might be easier to see if we
> consider two rockets traveling towards each other (their noses facing each
> other), and each wants to know the distance it must traverse to get from
> the nose of the other rocket to its tail. Then for example if each rocket
> is 10 meters long in its rest frame, and the two rockets have a relative
> velocity of 0.8c, each will measure only a 6 meter distance between the
> nose and tail of the other rocket, and the time they each measure to cross
> that distance is just (6 meters)/(0.8c).
>
> Jesse
>
>
>
>
> About mass, since the measured mass grows exponentially to infinity as v
> --> c, isn't this derivable from the LT, but in which frame? AG
>
>
> The notion of a variable relativistic mass is just an alternate way of
> talking about relativistic momentum, often modern textbooks talk solely
> about the latter and the only mass concept they use is the rest mass. For
> example the page at
> https://courses.lumenlearning.com/suny-physics/chapter/28-5-relativistic-momentum
> has a box titled "Misconception alert: relativistic mass and momentum"
> which says the following (note that they are using u to denote velocity):
>
> "The relativistically correct definition of momentum as p = γmu is
> sometimes taken to imply that mass varies with velocity: m_var = γm,
> particularly in older textbooks. However, note that m is the mass of the
> object as measured by a person at rest relative to the object. Thus, m is
> defined to be the rest mass, which could be measured at rest, perhaps using
> gravity. When a mass is moving relative to an observer, the only way that
> its mass can be determined is through collisions or other means in which
> momentum is involved. Since the mass of a moving object cannot be
> determined independently of momentum, the only meaningful mass is rest
> mass. Thus, when we use the term mass, assume it to be identical to rest
> mass."
>
> I'd say there's nothing strictly incorrect about defining a variable
> relativistic mass, it's just a cosmetically different formalism, but it may
> be that part of the reason it was mostly abandoned is because for people
> learning relativity it can lead to misconceptions that there is more to the
> concept than just a difference in how momentum is calculated, whereas in
> fact there is no application of relativistic mass that does not involve
> relativistic momentum. Momentum is needed for situations like collisions or
> particle creation/annihilation where there's a change in which objects have
> which individual momenta, but total momentum must be conserved. It's also
> used in the more general form of the relation of energy to rest mass m and
> relativistic momentum p, given by the equation E^2 = (mc^2)^2 + (pc)^2,
> which reduces to the more well-known E=mc^2 in the special case where p=0.
>
> By the way, since relativistic momentum is given by p=mv/sqrt(1 -
> v^2/c^2), you can substitute this into the above equation to get E^2 =
> (m^2)(c^4) + (m^2)(v^2)(c^2)/(1 - v^2/c^2), and then if you take the first
> term on the right hand side, (m^2)(c^4), and multiply it by (1 -
> v^2/c^2)/(1 - v^2/c^2) and gather terms, you get E^2 = [(m^2)(c^4) -
> (m^2)(v^2)(c^2) + (m^2)(v^2)(c^2)]/(1 - v^2/c^2), and two terms cancel each
> other out so this simplifies to E^2 = (m^2)(c^4)/(1 - v^2/c^2), and then if
> you take the square root of both sides you get E = γmc^2. So the original
> equation for energy as a function fo rest mass m and relativistic momentum
> p can be rewritten as E=Mc^2 where M is the relativistic mass defined as M
> = γm, again showing that relativistic mass is only useful for rewriting
> equations involving relativistic momentum.
>
> Jesse
>
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