Dear all
Few topics on Einstein I wrote sometime back. Now algebraically how
to prove it? Virtually the Einstein E=Mc^2, is the route through Newton's
laws of motion in 3 planes against the four dimensions of the Einstein.
Einstein formula and theory is still incomplete but it proves the advaita
philosophy of Potential energy of Sivam and Kinetic energy of shakti are
equal and one and the same. Now the first part shows scientific derivation
mathematically, but a bit tough to understand; the second part is simple ,
step by step maths algebra proving the einstein K Rajaram IRS 7923 Now,
*The "Proof" of Special Relativity*
When Einstein first proposed his Special Theory of Relativity
<https://www.emc2-explained.info/Special-Relativity/> in 1905 few people
understood it and even fewer believed it. It wasn't until 1919 that the
Special Theory was "proved by inference" from an experiment carried out on
his General Theory of Relativity. Physicists now routinely use relativity
in experiments all over the world every day of the year. However, many of
these experiments are highly specialised and usually require a great deal
of knowledge and training in order to understand them. So what evidence is
there for the general public? Probably the most spectacular "proof" is
nuclear weapons. These pages are not about the morality of such weapons
(but that's not to say that the question of their existence or use is not
an important one). However, whether one "likes" nuclear weapons or not no
one would deny that they exist. Both nuclear power and nuclear weapons are
derived from one principle; that mass can be converted into energy, and the
equation that exactly predicts that conversion is E = mc2. So what has that
to do with Special Relativity? The answer is that E = mc2 is derived *directly
*from Special Relativity. If relativity is wrong, then nuclear weapons
simply wouldn't work. Any theory or point of view that opposes Special
Relativity must explain where E = mc2 comes from if not relativity. Other
models of relativity that contain E = mc2 exist but here we are concerned
with the standard model as proposed by Einstein. This page explains, with
minimal mathematics, how E = mc2 is derived from Special Relativity. In
doing so it follows the same theoretical arguments that Einstein used.
However, for reasons of completeness, and although advanced and difficult
to follow, English translations of Einstein’s original Special Relativity
and E = mc2 papers are available here
<https://www.emc2-explained.info/Related-Pages/einsteins-papers.htm> (opens
in new tab).
*The Two Postulates*
The whole of Special Relativity is based on just two rules, or as they are
called in physics, postulates:
*Postulate I*: The principle of relativity:The laws of physics are the same
in all inertial frames.*Postulate II*: The principle of the constancy of
the speed of light:*The speed of light (in a vacuum) has the same constant
value c in all inertial frames.*
Jumping from these postulates to E = mc2 requires a little work. In order
to understand the following arguments it helps to be familiar with Special
Relativity, and in particular how moving at very high speeds dramatically
changes the properties of mass <https://www.si-units-explained.info/mass/>
and time <https://www.si-units-explained.info/time/>. If you aren't
familiar with these ideas you can read about the basics here
<https://www.emc2-explained.info/Special-Relativity/>.
*An Apparent Increase in Mass due to Speed*
One of the consequences of Special Relativity is that mass appears to
increase with speed. The faster an object goes, the "heavier" it seems to
get. This isn't noticeable in everyday life because the speeds we travel at
are far too small
<https://www.emc2-explained.info/Time-Dilation-at-Low-Speeds/> for the
changes to be apparent. In fact, an object needs to be moving at an
appreciable percentage of the speed of light (186,300 miles per second, or
300,000 kilometres <https://www.si-units-explained.info/length/> per
second) before any apparent mass increase starts to become noticeable in
everyday terms.
Where:
m = relativistic mass, i.e. mass <https://www.si-units-explained.info/mass/>
at the speed it is travelling.m0 = "rest mass", i.e. mass of object when
stationary.v = speed of object.c = speed of light.
If we examine a table of representative values for the speed of light
(below) we see that mass hardly increases at all until we reach about 75%
of c, but then starts to climb very rapidly.The second set of columns show
that beyond 99.9% of c the mass increase is very rapid indeed for even just
a very tiny increase in speed, tending towards (but never quite getting to)
infinity:
Note that the mass can never be smaller than unity (i.e.1). This may seem a
trivial point. After all, we can't just make the mass vanish into nothing.
However, while seemingly unimportant, we will return to this point later
and see that it is in fact essential to an understanding of how the
equation E = mc2 is derived.
Also note that the mass increase isn't felt by the object itself, just as
the time dilation <https://www.emc2-explained.info/Dilation-Calc/> of
Special Relativity isn't felt by the object. It's only apparent to an
external observer, hence it is "relative" and depends on the frame of
reference used. To an external observer it appears that the faster the
object moves the more energy is needed to move it. From this, an external,
stationary observer will infer that because mass is a resistance to
acceleration and the body is resisting being accelerated, the mass of the
object has increased.
*Kinetic Energy*
Next, we need to look at the energy involved in very high speed movement.
We have seen that as an object gets faster its mass appears to increase,
and the more mass an object has the more energy is required to move it. The
standard equation for the energy of movement (kinetic energy) is:
That is, kinetic energy is equal to half the mass multiplied by the
velocity squared. This is often called Newtonian kinetic energy. Note that
the velocity term is squared. This means, for example, that it takes far
more than twice the energy to travel at twice any particular speed. We can
see this by working through the equation for two values of v, where: v = 50
ms-1 and v = 100 ms-1 respectively, both with the same mass of 10 kg:
This equation is fine at "low" speeds, i.e. the speeds we encounter in
everyday life. However, we know that mass appears to increase as the speed
increases and so the Newtonian equation for kinetic energy must start to
become inaccurate at speeds comparable to the speed of light. So, how do we
compensate for the observed mass increase?
*Relativistic Kinetic Energy and Mass Increase*
In order to compensate for the apparent mass increase due to very high
speeds we have to build it into our equations. We know that the mass
increase can be accounted for by using the equation:
>From this equation we know that mass (m) and the speed of light (c) are
related in some way. What happens if we set the speed (v) to be very low?
Einstein realised that if this is done we can account for the mass increase
by using the term mc2 (although the exact arguments and mathematics
required to derive this are quite advanced and beyond the scope of these
pages). Using this term we now have an equation that takes into account
both the kinetic energy and the mass increase due to motion, at least for
low speeds:
This equation seems to solve the problem. We can now predict the energy of
a moving body *and* take into account the mass increase. What's more, we
can rearrange the equation to show that:
This result is fine for low speeds, but what about speeds closer to the
speed of light? We know that mass increases at high speeds, but according
to the Newtonian part of the equation that isn't the case. Therefore, we
need to replace the Newtonian part of the formula in order to make the
equation correct at *all* speeds. How can we do this?
We know that E – mc2 is approximately equal to the Newtonian kinetic energy
when v is small, so we can use E – mc2 *as *the definition of relativistic
kinetic energy:
We have now removed the Newtonian part of the equation. Note that we
haven’t given a formula for relativistic kinetic energy. The reason for
this will become apparent in a moment. Rearranging the result shows that:
It can now be seen that relativistic energy consists of two parts. The
first part is kinetic and depends on the speed of the moving body, while
the second part is due to the mass increase and does not depend on the
speed of the body. However, both parts must be a form of energy, but what
form? We can simplify the equation by setting the speed (i.e. the
relativistic kinetic energy) of the moving body to be zero, thereby
removing it from the equation:
We now have the famous equation in the form it's most often seen in, but
what does it mean?
We have seen that a moving body apparently increases in mass and has energy
by virtue of its speed (the kinetic energy). Looking at the problem another
way we can say that as the speed of a body gets lower there will be less
and less kinetic energy until at rest the body will have no kinetic energy
at all. So far so good, but what about the mass due to the speed of the
body? Again, as the body slows down the mass will become progressively
smaller but it *can't* reach zero. As noted near the start of the page, the
lowest the mass can be is unity (1) and we can't just make the body
disappear into nothing. The lowest possible mass the body can have is its
"rest mass", i.e. the mass the body has when it is at rest. But the
equation we have derived (E = mc2) isn't for mass, it's for *energy*. The
energy must somehow be locked up in the mass of the body.
Einstein therefore concluded that mass and energy are really different
manifestations of the same thing, i.e. that any mass is really tightly
packed energy. At the time he saw no mechanism for releasing the energy
from the mass, and was in fact sceptical
of the idea that it could ever be achieved. This didn't really matter to
Einstein, however. As a theoretical physicist he was happy that his
equations were consistent and he had a model to predict what happens to a
body moving at very high speeds. Converting matter to energy has of course
long since been achieved <https://www.emc2-explained.info/Emc2/Fission.htm>,
and you can see how much energy is released for a given amount of mass
using the E = mc2 calculator on this page
<https://www.emc2-explained.info/ECalc/index.htm>.
*The Full Form of the Equation*
So far we have referred to the energy of very high speeds as "relativistic
kinetic energy". This is fine for allowing us to work quickly through the
equations and to keep them simple, but there has to be a more formal way of
expressing what we mean. As with the term mc2, the detailed derivation of
the full equation for E = mc2 is quite complex. However, to those familiar
with the basic mathematics of Special Relativity
<https://www.emc2-explained.info/Special-Relativity/> the way in which we
take into account the kinetic energy of E = mc2 will come as no surprise:
For any non-stationary body the total energy is given as:
This equation takes into account the total energy (E), the mass of the body
(m), and the speed of the body (v). As such it accounts for both the
relativistic mass increase and the relativistic kinetic energy.
*Conclusion*
The subject of this page is quite difficult to understand, even though much
of the more difficult mathematical considerations have been left to one
side. The conceptual jump from the two postulates of Special Relativity to
the equivalence of mass and energy is certainly not obvious, and it's
extraordinary that Einstein proposed it long before there were any
experimental results to indicate the true nature of the relationship
between mass and energy.
The results of Einstein's work in this area are far more widespread than is
usually thought and affect everyone on the planet. As with all science we
can use the results for "good" and "bad". The reason the words "good" and
"bad" are in quotes is because it depends on your point of view. For
example, you may think that nuclear power stations (which use E = mc2
directly) are either a good or a bad thing. Likewise, depending on your
point of view, nuclear weapons are either a good or a bad thing; they
either ended one war and prevented another, or they are immoral and bound
to fall into the wrong hands sooner or later. On the other hand, there have
been great advances in using E = mc2 in the medical field, particularly to
treat cancer. Again though, this can be seen as either a good thing (curing
a disease) or a bad thing (overburdening an already over-populated planet).
Although these issues are undoubtedly important it's not for pages such as
this to hold an opinion either way, but to merely explain some of the
science behind them. It's far too late to "un-invent" E = mc2 and the best
we can do is to use it in an informed way for the things that we believe
are worthwhile
I Acceleration = (Final Velocity-Initial Velocity) / Time Taken
a = v-u /t or at = v-u
v = u + at 1 law of motion one
*II Derivation of Second Equation of Motion*
- *Simple Algebraic Method:*
Let the distance be “s”.
Distance = Average velocity × Time. Also, Average velocity (u+v)/2
Distance (s) = (u+v)/2 × t
Also, from v = u + at
s = (u+u+at)/2 × t = (2u+at)/2 × t
s = (2ut+at²)/2 = 2ut/2 + at²/2
or s = ut +½ at²
*III Derivation of Third Equation of Motion*
- *Simple Algebraic Method*
We have, v = u + at. Hence, we can write t = (v-u)/a
Also, we know that, Distance = average velocity × Time
Therefore, for constant acceleration we can write: Average velocity
<http://toppr.com/guides/physics/motion-in-a-straight-line/average-velocity-and-average-speed/>
=
(final velocity + initial velocty)/2 = (v+u)/2
Hence, Distance (s) = [(v+u)/2] × [(v-u)/a]
or s = (v² – u²)/2a
or 2as = v² – u²
or v² = u² + 2as 3rdlaw of motion
m=m0 /^1-v2 / c2
How to prove m=m0√1−(v2c2)?
Let us consider two inertial frames of reference S and S′. Let S is a
stationary frame of reference with a stationary observer in it and S′ is
the moving frame of reference. In the beginning, suppose these are at the
same position so that respective observers O and O′ coincide. After
time t=t′=0 frame S′ begin to move with a uniform velocity v along x-axis.
Suppose there are two particles moving towards each other in the
frame S′. According to the observer O′, velocity of particle A will
be u′ and of particle B will be –u′ .
As observed from frame S, Velocity of A is u1 and B is u2. These velocities
are given by relativistic addition of velocity expressions as below
u1=u′+v1+u′vc2 .............(1) m1 v1 +m2v2 conservation of momentum
u2=−u′+v1−u′vc2 ..............(2)
Let m1andm2 be respective masses of A and B as observed from frame S.
Let the particles collide with each at a certain instant and momentarily
come to rest. Even while at rest, these travel with the velocity v of the
frame S′.
According to the Law of conservation of momentum
<https://socratic.org/physics/momentum-and-collisions/conservation-of-momentum>
we
have S AND s’ being in the opposite reaction + and minus or else it ie +
and + only.
m1u1+m2u2=(m1+m2)v=m1v+m2v ……………………………….(.3)
transforming yellow and Red in the equation, 3
[m1u1-m1v=m2v-m2u2]
*⇒**m1(u1–v)=m2(−u2+v)…*…………………………4
Using equations (1) and (2) in above we get
m1⎡⎣u′+v1+u′vc2–v⎤⎦=m2⎡⎣v−−u′+v1−u′vc2⎤⎦
*m1**⎡⎢* ⎢⎣{u′(+v−v)} {=0](1+u′vc2),,,,,,,,,,,,,,,,,,,,,,,5 hence
*1+u′vc2**⎤⎥………………**6 m1(1+u’vc2//*
m2=--{--u’+v1-u’vc2}+v (substituted)
=u’-v1+u’vc2+v
*⎥⎦**=m2**⎡⎢* ⎢⎣v(1−u′vc2)+u′−v1−u′vc2⎤⎥⎥⎦
⇒m1⎡⎣u′−u′v2c21+u′vc2⎤⎦=m2⎡⎢⎣−u′v2c2+u′1−u′vc2⎤⎥⎦
⇒m1⎡⎣11+u′vc2⎤⎦=m2⎡⎣11−u′vc2⎤⎦
⇒m1m2=1+u′vc21−u′vc2 .............(3)
Now writing LHS of (1) as
1−u21c2
Similarly RHS can be written as
1−(u′+v1+u′vc2)2c2
⇒1−⎛⎝u′+vc1+u′vc2⎞⎠2
⇒(1+u′vc2)2−(u′+vc)2(1+u′vc2)2
Simplifying and equating with rewritten (1) we get
1−u21c2=1+(u′v)2c4−(u′)2c2−v2c2(1+u′vc2)2 ................(4)
Similarly we get from (2)
1−u22c2=1+(u′v)2c4−(u′)2c2−v2c2(1−u′vc2)2 ............(5)
On dividing equation (5) by (4) and taking square root of both sides we get
√1−u22c2√1−u21c2=1+u′vc21−u′vc2 .............(6)
Comparing (3) and (6) we get
*m1m2=√1−u22c2√1−u21c2 .......(7)*
*Let us assume that the particle B is at rest as seen from frame S that is
it has zero velocity before collision*
*⇒**u2=0*
*also writing m2=m0 where m0 is the rest mass of this particle, With
this **equation
(7) reduces to*
*m1m0=1√1−u21c2*
*Using common notation as m1=mandu1=v, above equation becomes*
*mm0=1√1–v2c2 **⇒*
*m=m0√1–v2c2 .......(8) This is the required derivation.*
*THE ABOVE IS DERIVATION WHICH CAN BE UNDERSTOOD IF MATHS AND PHYSICS ARE
STILL REMEMBERED. or else*
*There is another way of deriving the same in a short cut method which is
Narrated Below: *
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
1 E=mc^2 and E=1/2mv^2 are both equations that relate to energy, but
they describe different types of energy. The equation E=mc^2 relates to the
concept of mass-energy equivalence, which means that mass and energy are
interchangeable. This equation states that the amount of energy (E) that an
object has is equal to its mass (m) multiplied by the speed of light (c)
squared. On the other hand, the equation E=1/2mv^2 relates to the kinetic
energy of an object in motion. This equation states that the amount of
kinetic energy (E) that an object has is equal to half of its mass (m)
multiplied by its velocity (v) squared. So while both equations relate to
energy, they describe different types of energy and have different
applications.
2 In the equation [math]e=\frac{1}{2} mv^2[/math], the
[math]\frac{1}{2}[/math] is a coefficient of the kinetic energy,
[math]e[/math], which represents the amount of energy an object possesses
due to its motion. On the other hand, in the equation [math]E=mc^2[/math],
the [math]\frac{1}{2}[/math] is not present. This equation represents the
relationship between the rest energy, [math]E[/math], and the mass,
[math]m[/math], of an object at rest, where [math]c[/math] is the speed of
light. So, to answer your question, the [math]\frac{1}{2}[/math] is not
present in the equation [math]E=mc^2[/math].
3 When comparing e=1/2 mv^2 and E=mc^2 where does the 1/2 go?
E=mc2= is the relativistic ‘total’ energy of a particle, including its rest
energy m0c^2.
4 When we write out the relativistic mass, this total energy is
E=m01sq root of v^2/c^2*c^2
e=1/2m0v^2=is the non-relativistic kinetic energy. Note that here m stands
for the rest mass.
5 In relativity theory the kinetic energy of a particle is given as the
total energy minus its rest energy. == mc^2-1/2mv^2 ( Einstein minus
newton)
The formula for speed is speed = distance ÷ time. To work out what the
units are for speed, you need to know the units for distance and time. In
this example, distance is in metres (m) and time is in seconds (s), so the
units will be in metres per second (m/s).
1 v=u+at 1 law of newton
2 v² = u² + 2as. 2 law F=ma
3 s = ut + ½at² 3rd law
6 Ekin=m01+v2c2−−−−−√c2−m0c2=m0c2[1+v2c2−−−−−√−1]
Ekin=m0 sq root of 1+ v^2/c^2*c^2 - m 0 c^2
= m0c^2 {√(1+v^2/C^2 ) -1}
√(1+v^2/c^2 ⋅)=
{1+v^2/c^2}1/2-1
= 11/2 +v^2^1/2//c^2^1/2 -1
= 1-1+1/2*v^2/c^2
= 0+ ½* v^2/c^2 = 1/2×ν^2/c^2
If you evaluate the square brackets for velocities v that are much smaller
than c, you will find that
= 1/2×ν^2/c^2
Insert that into the kinetic energy and you will get
Ekin = m0c2*1/2*v^2/c^2
= m0 1/2v^2
7 E_k≔1/2 m_0 V_^2 this is Newton energy.
Now in space momentum m1 is
Ek1 =
E_k≔1/2 m_0 V_^2
8 NOW both in motion plus momentum=
E_k≔1/2 m_0 V_^2 +(in motion is newton) -----------1
Ek1 = 1/2 m_0 V_^2 ---in momnum ---2
E_k + Ek1------------= E ------------------------------------2A
1/2 m_0 V_^2 + 1/2 m_0 V_^2 ------= m0v^2---------2B
where E- Total energy; m is Mass and V is speed of light in Einstein and
speed less than light in Newton and 4 dimension in Einstein and 3 dimension
in Newton.
= E=Mc^2 =Einstein----------------------------------------3
KR IRS 7923
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