Relativistic Kinetic Energy

  • #1

Main Question or Discussion Point

I am reading "Relativity" by Einstein right now, and I came across a formula he gave for kinetic energy in accordance with his theory of relativity. He says that "the kinetic energy of a material point of mass m is no longer given by the well-known expression

.5m(v^2)

but by the expression

[m(c^2)]/[sqrt(1-(v^2)/(c^2))]

I'm assuming this expression was found by the Lorentz transformation, but he doesn't show that. He goes on to explain how this formula suggests the impossibility of speeds faster than c, since the fraction approaches infinity as v approaches c.

The reason why I am confused is that the two expressions are extremely different. If the mass is 2 kg and the velocity is 3 m/s, then by the first expression, the KE is 9 J. By the second expression, you get about
180,000,000,000,000,000 J. What am I missing?

(also, what is a material point?)

Thanks for any help on this
 
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Answers and Replies

  • #2
I am reading "Relativity" by Einstein right now, and I came across a formula he gave for kinetic energy in accordance with his theory of relativity. He says that "the kinetic energy of a material point of mass m is no longer given by the well-known expression

.5m(v^2)

but by the expression

[m(c^2)]/[sqrt(1-(v^2)/(c^2)]

I'm assuming this expression was found by the Lorentz transformation, but he doesn't show that. He goes on to explain how this formula suggests the impossibility of speeds faster than c, since the fraction approaches infinity as v approaches c.

The reason why I am confused is that the two expressions are extremely different. If the mass is 2 kg and the velocity is 3 m/s, then by the first expression, the KE is 9 J. By the second expression, you get about
180,000,000,000,000,000 J. What am I missing?

(also, what is a material point?)

Thanks for any help on this
The expression you propose is not correct.
It is if I remember well it is
K=mcc[(1-vv/cc)^-1/2 -1]
 
  • #3
If the expression I gave above is incorrect then there is a typo in the book because that is exactly what the book has in it - I double-checked.
 
  • #4
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The form given in the book includes the rest energy term [itex]mc^2[/itex].

If you know how to obtain the Taylor series for a function, try this: expand the form given for the energy in powers of v/c, and then subtract off the rest energy term. Take a look at what you have left, and remember that for non relativistic speeds, v/c is usually small, and so higher powers of v/c are even smaller.
 
  • #5
Actually the kinetic energy of a body is given as

[tex]K.E=(\gamma - 1)m_0c^2[/tex]
 
  • #6
Ke

Actually the kinetic energy of a body is given as

[tex]K.E=(\gamma - 1)m_0c^2[/tex]
Compring my answers with yours some people will blame me for using m for the rest mass and you for using m(0). That is the situation when we do not use all the same notation for the same phyhsical quantity. I like more m(0).:rofl:
 
  • #7
isn't [tex]m_0[/tex] standard notation????
 
  • #8
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0
isn't [tex]m_0[/tex] standard notation????
There really is no standard notation for mass. Workers in different fields use different symbols according to the objective of their work. There's even a paper by Einstein which was published in 1907 in which he used neither for proper mass (aka rest mass). He used [itex]\mu[/itex]. This is what I prefer myself. The invariants in relativity, i.e. the proper quantities, often have the greek symbol on place of the latin symbol. So instead of s for distance the symbol [itex]\sigma[/itex] is used. Instead of t for time the symbol [itex]\tau[/itex] is used. Thus the extention of this idea to "m". Instead of m the symbol [itex]\mu[/itex] is used.

Pete
 
  • #9
robphy
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The invariants in relativity, i.e. the proper quantities, often have the greek symbol on place of the latin symbol. So instead of s for distance the symbol [itex]\sigma[/itex] is used. Instead of t for time the symbol [itex]\tau[/itex] is used. Thus the extention of this idea to "m". Instead of m the symbol [itex]\mu[/itex] is used.

Pete
Unfortunately, this idea does not extend to [tex]\beta[/tex] and [tex]\gamma[/tex] [unless one interprets these quantities as invariants that depend on two 4-velocities].
 
  • #10
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5
There really is no standard notation for mass. Workers in different fields use different symbols according to the objective of their work. There's even a paper by Einstein which was published in 1907 in which he used neither for proper mass (aka rest mass).
I prefer rest mass over relativistic mass because relativistic mass is not Lorentz invariant.

But I realize that, in the macro world, relativistic mass is used more often than rest mass.

For example measuring the rest mass of a spaceship even when it is at rest is not as simple as it appears to be. For instance if we were to use a balance we would have to take into account the composition of the material on each side of the balance.
 
  • #11
mass on the balance

I prefer rest mass over relativistic mass because relativistic mass is not Lorentz invariant.

But I realize that, in the macro world, relativistic mass is used more often than rest mass.

For example measuring the rest mass of a spaceship even when it is at rest is not as simple as it appears to be. For instance if we were to use a balance we would have to take into account the composition of the material on each side of the balance.
It is not enough that the masses on the pans of the balance equate each other? Please explain.
 
  • #12
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5
It is not enough that the masses on the pans of the balance equate each other? Please explain.
The restmass and relativistic mass differential depends on the atomic numbers of the elements of the composition.

Take for instance the mass of a gold bar, the restmass and relativistic mass are not identical even when that gold bar does not appear to be moving. The electrons do move and in the case of gold some of them reach relativistic speeds.

Obviously, the difference is very small but nevertheless it is not zero!
 
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  • #13
I think we're kind of getting off topic here. So, the reason why the book's expression for relativistic KE gives us a value 20 quadrillion times greater in my example than the non-relativistic KE is not that the expression in the book is incorrect, but that it includes the rest energy? Why should this account for such a huge difference though?
 
  • #14
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I think we're kind of getting off topic here. So, the reason why the book's expression for relativistic KE gives us a value 20 quadrillion times greater in my example than the non-relativistic KE is not that the expression in the book is incorrect, but that it includes the rest energy?
Yes.

Why should this account for such a huge difference though?
Because [itex]m_0 c^2 >> m_0 v^2[/itex] at non relativistic speeds!
 
  • #15
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I prefer rest mass over relativistic mass because relativistic mass is not Lorentz invariant.
I don't follow you. Would you prefer proper time over time because the former is a Lorentz invariant? Would you prefer proper length over length? Would you prefer proper energy over energy? What does it mean for you to prefer one over the other??

Pete
 
  • #16
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Unfortunately, this idea does not extend to [tex]\beta[/tex] and [tex]\gamma[/tex] [unless one interprets these quantities as invariants that depend on two 4-velocities].
Neither of those are intrinsic properties of a body so it really doesn't apply here. I think you know where I was going with what I said anyway.

Pete
 
  • #17
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I don't follow you. Would you prefer proper time over time because the former is a Lorentz invariant? Would you prefer proper length over length? Would you prefer proper energy over energy? What does it mean for you to prefer one over the other??

Pete
I'm not sure what MeJennifer meant (of course), but I'd just like to add that if I made a similar statement, I would mean that in situations where I can use either notions of mass correctly, I would prefer to use the rest mass.

My reasoning would be as follows: wherever I can I would prefer to use tensorial quantities because to me it feels like physical systems/properties can be represented best by tensors on spacetime. I know that in certain situations using non-tensors can simplify matters, and even in some cases where we are forced to consider non-tensors (e.g. Christoffel symbols), but if I can, I'd use the tensors.
 
  • #18
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I'm not sure what MeJennifer meant (of course), but I'd just like to add that if I made a similar statement, I would mean that in situations where I can use either notions of mass correctly, I would prefer to use the rest mass.
In a situation in which you could use both then if you were in a position to use Energy or proper energy which would you choose?

Do you see where I'm going here? These questions/selections are far to vauge to generalities and one is never replaceable with the other, ... , at least I can't think of one. How you you MJ and masudr?

Pete
 
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  • #19
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In a situation in which you could use both then if you were in a position to use Energy or proper energy which would you choose?

Pete
I'm not entirely sure of the difference... as far as I know energy is the zeroth component of the energy-momentum 4-vector

[tex]p^\mu = m_0 \frac{dx^\mu}{d\tau}[/tex]

So what do you mean by energy and proper energy? As in, what's the difference between them?
 
  • #20
2,946
0
I am reading "Relativity" by Einstein right now, and I came across a formula he gave for kinetic energy in accordance with his theory of relativity. He says that "the kinetic energy of a material point of mass m is no longer given by the well-known expression

.5m(v^2)

but by the expression

[m(c^2)]/[sqrt(1-(v^2)/(c^2))]

I'm assuming this expression was found by the Lorentz transformation, but he doesn't show that. He goes on to explain how this formula suggests the impossibility of speeds faster than c, since the fraction approaches infinity as v approaches c.

The reason why I am confused is that the two expressions are extremely different. If the mass is 2 kg and the velocity is 3 m/s, then by the first expression, the KE is 9 J. By the second expression, you get about
180,000,000,000,000,000 J. What am I missing?

(also, what is a material point?)

Thanks for any help on this
You are quite correct in what Einstein's book says on this. Consider this: If the rest energy of the electron is constant then it can be dropped from the equation for the conservation of energy leaving only [itex]\gamma[/itex]m0c2 which is what Einstein has in his book. The meaningful quantities now are the changes in kinetic energy.

Hope that helps

Pete
 
  • #21
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0
I'm not entirely sure of the difference... as far as I know energy is the zeroth component of the energy-momentum 4-vector

[tex]p^\mu = m_0 \frac{dx^\mu}{d\tau}[/tex]

So what do you mean by energy and proper energy? As in, what's the difference between them?
Energy is proportional to the time component of 4-momentum whereas rest energy is proportional to the magnitude of this vector. Each has a significan meaning and one cannot be exchanged with the other.

Can you give me an example from what you said in your previous post? I.e. ..in situations where I can use either notions of mass correctly, I would prefer to use the rest mass. A solid example will help me better understand what you mean. What I see right now is no such examples exist since they would be addressing different questions/topics.

Thanks masudr

Pete
 
  • #22
1,997
5
I don't follow you. Would you prefer proper time over time because the former is a Lorentz invariant? Would you prefer proper length over length? Would you prefer proper energy over energy? What does it mean for you to prefer one over the other??
I think there are two ways to approach the intricacies of relativity. You can approach it from one single 4-dimensional space-time reality or from a set of hyper surfaces with constant proper time.

If you take the first view, you immediately see that it only makes sense to think in terms of Lorentz invariant properties. You only work with concepts that are space-time related, 4-vectors like 4-velocity, 4-acceleration, energy-momentum, as well as proper time and rest mass.

In the second case, you work with values that are not valid for space-tine as a whole, but only valid for those observers whose proper time is orthogonal to the hyper surface. All the values are different for an observer who is moving relative to you by the Lorentz factor.

To me, Lorentz variant properties are incomplete perspectives of Lorentz invariant properties.
 
  • #23
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I think there are two ways to approach the intricacies of relativity. You can approach it from one single 4-dimensional space-time reality or from a set of hyper surfaces with constant proper time.

If you take the first view, you immediately see that it only makes sense to think in terms of Lorentz invariant properties. You only work with concepts that are space-time related, 4-vectors like 4-velocity, 4-acceleration, energy-momentum, as well as proper time and rest mass.

In the second case, you work with values that are not valid for space-tine as a whole, but only valid for those observers whose proper time is orthogonal to the hyper surface. All the values are different for an observer who is moving relative to you by the Lorentz factor.

To me, Lorentz variant properties are incomplete perspectives of Lorentz invariant properties.
Eventually you have to take measurements or define a tensor or vector or 1-form and when you do this you're looking at non-invariant properties. E.g. the properties of inertia of a body is nicely described by rel-mass. When we take measurements of length then proper lengths are not always useful. E.g. its much easier to understand relativistic electrodynamics when you look at the components of 4-vectors since in some cases the magnitude of the invariant quantity tells you nothing about the system. E.g. consider the charge density of a current carrying wire where in the rest frame S the wire is electrically neutral. Analyzing this situation with invariant values is not helpful. See the web page I made on this point at

http://www.geocities.com/physics_world/em/rotating_magnet.htm

Best wishes MJ

Pete
 
  • #24
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Energy is proportional to the time component of 4-momentum whereas rest energy is proportional to the magnitude of this vector. Each has a significan meaning and one cannot be exchanged with the other.
So what I have called energy is the time component of 4-momentum: [itex]\gamma m_0 c^2[/itex], and what you have called proper energy, and are now calling rest energy is proportional to the length of the 4-momentum which is [itex]m_0^2 c^2[/itex]. In your post #18, you ask me which I would use given the freedom to do so: I don't see any situation where I have this freedom.

What I meant was that (and I must admit my use of special relativity has only been relevant to particles) if I had the freedom to use relativistic mass [itex]m=\gamma m_0[/itex] and rest mass [itex]m_0[/itex] I would rather use the rest mass and explicitly put in [itex]\gamma[/itex] as and when required.
 
  • #25
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So what I have called energy is the time component of 4-momentum: [itex]\gamma m_0 c^2[/itex], and what you have called proper energy, and are now calling rest energy is proportional to the length of the 4-momentum which is [itex]m_0^2 c^2[/itex]. In your post #18, you ask me which I would use given the freedom to do so: I don't see any situation where I have this freedom.

What I meant was that (and I must admit my use of special relativity has only been relevant to particles) if I had the freedom to use relativistic mass [itex]m=\gamma m_0[/itex] and rest mass [itex]m_0[/itex] I would rather use the rest mass and explicitly put in [itex]\gamma[/itex] as and when required.
That's fair enough. :approve:

However let us put your idea to the test. Place an ion which otherwise has a proper mass of "m" (I'm getting too lazy to type in the subscripts) in a linear particle accelerator. There is an electric field parallel to the accelerator tube and as such the ion will be in an electric field (make it uniform for simplicity). Now what is the "rest mass" of this ion?

Pete
 
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