# Energy-momentum equation

1. Feb 3, 2010

### dimsun

Restmass:
$$m_0 = \sqrt{\frac{E^2}{c^4} - \frac{p^2_x}{c^2} - \frac{p^2_y}{c^2} - \frac{p^2_z}{c^2}}$$

Relativistic mass:
$$m_{rel} = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}$$

$$\sqrt{m_{rel}^2 - \frac{m_{rel}v^2_x}{c^2} -\frac{m_{rel}v^2_y}{c^2} -\frac{m_{rel}v^2_z}{c^2}} = \sqrt{\frac{E^2}{c^4} - \frac{p^2_x}{c^2} - \frac{p^2_y}{c^2} - \frac{p^2_z}{c^2}}$$

So it seems that the energy-momentum equation must be extended:
$$m^2 - s^2_xc^2 - s^2_yc^2 - s^2_zc^2 = \frac{E^2}{c^4} - \frac{p^2_x}{c^2} - \frac{p^2_y}{c^2} - \frac{p^2_z}{c^2}$$

And that we have to explore the new quantity 's'.
Because of periodicity we don't have to extend the number of dimensions to more then 8.
We can describe this relationship by means of two opposite quaternions.

2. Feb 4, 2010

### torquil

The energy-momentum equation relates rest mass m0, energy, and momentum.

You have contructed an equation relating relativistic mass, energy, momentum and velocity. Btw, you have forgotten to square all the m_rel's inside your left square root.

Actually what you have done is akin to writing "0=0". Remember that p = gamma*m0*v where gamma is the well-known 1/sqrt(1-v^2/c^), m0 is the rest mass, and v is the velocity. Also, Einstein's equation E = mc^2 includes the relativistic mass. Put those in under your left square root, and you have in effect written:

sqrt(E^2 - p^2) = sqrt(E^2 - p^2), ie similar to 0=0

So this does not contradict the usual energy-momentum relation. Actually this 0=0 equation appears because you have used circular logic.

Torquil

Last edited: Feb 4, 2010
3. Feb 4, 2010

### Meir Achuz

Your conundrum illustrates the danger of trying to still use the concept of
"relativistic mass" in special relativity.

4. Feb 5, 2010

### dimsun

There are more subtilities involved.
The concept of mass isn't cristalized at all.
Meir Achuz, you called it "the danger of trying to still use the concept of 'relativistic mass' in special relativity".
And Einstein himself: "It is not good to introduce the concept of the mass of a moving body for which no clear definition can be given. It is better to introduce no other mass concept than the ’rest mass’ m. Instead of introducing M it is better to mention the expression for the momentum and energy of a body in motion."
And that is what is done: A new quantity s instead of relativistic mass.

What is the relationship between restmass, relativistic mass, invariant mass, proper mass and coordinate mass?
For example:
$$\tau =$$ Proper time and t = coordinate time:

$$\tau = \sqrt{t^2 - \frac{l_x^2}{c^2} - \frac{l_y^2}{c^2} - \frac{l_z^2}{c^2}}$$

Makes this sense if we know that restmass:

$$mc^2 = \sqrt{ E^2 - p_x^2c^2 - p_y^2c^2 - p_z^2c^2}$$

It would make more sense if this relationship isn't restmass but is proper energy:

$$E_p = \sqrt{E - p_x^2c^2 - p_y^2c^2 - p_z^2c^2}$$
(Torquil in your line of reasoning $$E = mc^2$$ is also 0 = 0)

Now the subtility is in this that mass and energy are different quantities. Some particles have energy but no mass. Einstein said that energy and mass are different manifestations of the same thing.

By the way, Minkowski invented spacetime. In the Minkowski metric, time t is replaced by the proportional imaginairy quantity ict. If we apply this to the mass energy relation then we get: $$E = m(ic)^2$$ therefore $$E = -mc^2$$. If we don't add this minus sign, then the mass-energy equivalence isn't consistent with special relativity.

We have the equivalences:
$$E = \frac{mc^2}{\sqrt{1 - \frac{v^2}{c^2}}}$$

$$p = \frac{m_0v}{\sqrt{1 - \frac{v^2}{c^2}}}$$

$$m_{rel} = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}$$

The new quantity s is suggested in the following equivalence:

$$s = \frac{\frac{m}{v}}{\sqrt{1 - \frac{v^2}{c^2}}}$$

In which v is the velocity of the reference frame.

5. Feb 5, 2010

### torquil

I disagree with most of your statements. Be careful to only use one type of mass so that you donæt mix them up in your calculations. I prefer to just use m for restmass, and never use relativistic mass. There is no need for it, since it is related to r by multiplication with gamma. So best to leave it out alltogether.

Also, it is important to specify the initial independent assumptions clearly. Any good textbook in SR will do this precisely. The first consequences of those will never be "0=0" statements. If you equate results that are both consequences of the same assumptions, they have the potential of reducing to 0=0. You could e.g. compare your results with this:
http://galileo.phys.virginia.edu/classes/252/energy_p_reln.html

Unfortunately, they also use both rest and relativistic mass, even though it is not necessary.

"Relativistic mass" is simply the same as gamma(v)*m where m is restmass, so it is not necessary to introduce it, because sometimes it creates confusion, particularly when forgetting to specify which masses that belong in which formulae.

What you say about the consequences of the "imaginary time" interpretation of Minkowski space is not correct. You cannot suddenly replace c by ic in a formula that is already derived for Minkowski space. The quantity 'c' is already defined as speed in Minkowski space, so there is no need to change to imaginary time there.

The "imaginary time" trick is used when going from equations defined on euklidean space to minkowski space. For your formulas in the previous post, that has already been done, so there is no need to do it again, becaues then you end up back in euklidean space! :-)

Torquil

6. Feb 6, 2010

### dimsun

Torquil, I like your attacks. All new physical models need attack in order to see if they will survive.

The consequences of the imaginary units in Minkowski spacetime are correct. It is not 'suddenly' replacing c by ic. It is all about the difference between non-relativistic quantities and relativistic quantities. In non-relativistic mechanics we can use time t and lenth l. But that changes when we use relativistic mechanics. In relativistic mechanics all quantities are connected to each other. In that case the quantities are time ict and length l, or we also can use time t and length -i l/c. (I prefer this last description because in the quaternion picture we have three imaginairy units that we can apply to the three length dimensions.)

According to Einstein the discovery Minkowski did, was important for the formal development of the theory of relativity. The importance lays down in the fact that he saw that the 4 dimensional spacetime continuum of the special theory of relativity in his most essential formal property had a very pronounced relationship with the 3 dimensional continuum of euclidian geometry. To make this relationship more clear we have to replace the usual time coordinate t with the proportional imaginary quantity ict. Then the laws of physics, (that satisfy the demands of special relativity) take the mathematical form in which the time coordinate exactly plays the same rol as the three space coordinates.

In case of the mass-energy equavalence, I haven't seen anyone to include the consequences of special relativity that clearly demands that E = m(ic)2

7. Feb 6, 2010

### torquil

Euklidean 4-spacetime is

$$\mathbb{E}^4:={(t',\mathbf{x}') \in \mathbb{R}^4$$

with square line element

$$ds'^2 = dt'^2 + d\mathbf{x}'^2$$

This spacetime doesn't describe our observed universe. In essence, it is an abstract mathematical construction. The speed of light is not described by $$d\mathbf{x}'/dt'$$.

Minkowski 4-spacetime is

$$\mathbb{M}_4:={(t,\mathbf{x}) \in \mathbb{R}^4$$

with square line element

$$ds^2 = -dt^2 + d\mathbf{x}^2$$

As you see from the last two equations, both $$t$$ and $$\mathbf{x}$$ are real. The speed of light is described in terms of these real variables, i.e. c=dx/dt where dx and dt are infinitesimal increments along the light's trajectory which satisfies ds^2=0. If you replace $$t$$ by $$it$$ here in the expression for c, you end up with the velocity of an unphysical trajectory in Euklidean space, unfortunately.

In itself, the fact that Minkowski spacetime can be described by letting the $$t$$-coordinate in Euklidean spacetime become imaginary doesn't mean anything physical. It is only a computational convenience in some situations.

Here is a good free reference for special relativity (p.69 and onwards):
http://www.uio.no/studier/emner/matnat/fys/FYS3120/v10/undervisningsmateriale/LectureNotes3120.pdf [Broken]

Torquil

Last edited by a moderator: May 4, 2017
8. Feb 6, 2010

### Altabeh

Of course using imaginary time is a way to make more detailed equations simple to deal with! For instance, you can put

$$x'=at+bx$$ and
$$t'=ct+dx$$,

into the equation

$$d{s'}^2=ds^2$$,

where $$ds^2$$ shows the Minkowski line-element, to get the Lorentz boost. But this can be a little bit cumbersome and boring so physicists prefer to use imaginary time scales

$$T=ict,\ T'=ict'$$

to avoid writing too much and this will give the same result as the first approach! but this is totally a matter of convetion which depends upon the author's taste to be made use of!

AB

9. Feb 7, 2010

### dimsun

I do not understand, why is this trajectory unphysical according to you?

In relativistic mechanics velocity get the following expression:

$$-i \frac{\mathbf{v}}{\mathbf{c}} = \frac{\Delta \mathbf{x}}{\Delta\mathbf{ict}}$$

This term pops up in the Lorenz factor.

And the speed of light becomes totaly imaginair:

$$-i = \frac{d \mathbf{x}}{d \mathbf{ict}}$$

It is not that we can write time t as ict and length l, but the value of doing this lays in the whole structure of the (hyper)complex numbers. The structure of the (hyper)complex numbers describes the structure of spacetime much better then the real numbers do. The structure of (hyper)complex numbers describes the relation between the relativistic quantities. A physical model is not a description of reallity, but physical models is to describe how physical phenomena work. So computational convenience is a good thing.

Dimsun

10. Feb 7, 2010

### kg4pae

That is an odd way to write that equation. Oh well...
Forget relativistic mass. It's an obsolete concept that even Einstein hated. Mass is considered a scalar, meaning it has the same value in all reference frames: like the speed of light in a vacuum.
I don't see why.
However, quaternions form only a division ring and not a field. By using the traditional method, we have something like a field structure for events.

11. Feb 7, 2010

### kg4pae

Yes, however...
Multiply this equation through by i and get

$$1=i\frac{d x}{d ict}=\frac{1}{c}\cdot\frac{d x}{d t}=\frac{v}{c}$$

Thus v=c, which is not imaginary, just too restrictive.
I think you're just making it too complicated. That is probably why you are introducing self-inconsistencies in your arguments. Keep it simple and usually it will work out better.

73s and clear skies.

12. Feb 7, 2010

### torquil

The "imaginary-time" viewpoint of SR is fine, and it is consistent. The name however is confusing, because time is not actually imaginary in this formalism. The last component of the four vector is imaginary, but also represents $$ict$$, not t. The $$c$$ and $$t$$ are still real. It doesn't lead to E=-mc^2, or a modification of the energy-momentum relation, because it is easily seen to be equivalent with the more powerful pseudo-Riemannian approach to SR.

All speeds are real numbers in this formalism. Using imaginary time, the trajectory of light consists of four-vectors of the form

$$(x,y,z,ict)$$ where $$x,y,z,c,t\in\mathbb{R}$$

All of these position/speed/time parameters are real.

The vector that encodes this data has an $$i$$ in its last component, so the vector itself is an element in a metric vector space where one axis consists of imaginary numbers, in order to get the correct inner product between pairs of vectors. Velocities for trajectories are still defined with appropriate factors if $$i$$ so that they come out as real numbers in the end.

Many professionals of SR and GR nowadays discourage the use of the imaginary time formalism, because it makes the transistion to the more general geometries of GR harder for the students. I think it is better to choose a formalism that can be considered a special case of GR.

Torquil

13. Feb 8, 2010

### dimsun

Relativistic mass is not well understood, but it is mathematically sound.

Relativistic energy:
$$E_{rel} = \frac{E_0}{\sqrt{1 - \frac{v^2}{c^2}}}$$

$$E_0 = \sqrt{E_{rel}^2 - \frac{E_{rel}^2v_x^2}{c^2} - \frac{E_{rel}^2v_y^2}{c^2} - \frac{E_{rel}^2v_z^2}{c^2}$$

$$E_0 = \sqrt{E^2 - p_x^2c^2 - p_y^2c^2 - p_z^2c^2}$$

In which p is momentum ofcourse.

Relativistic mass:
$$m_{rel} = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}$$

$$m_0 = \sqrt{m_{rel}^2 - \frac{m_{rel}^2v_x^2}{c^2} - \frac{m_{rel}^2v_y^2}{c^2} - \frac{m_{rel}^2v_z^2}{c^2}$$

$$m_0 = \sqrt{m^2 - s_x^2c^2 - s_y^2c^2 - s_z^2c^2$$

Hate is not a good scientific reason to reject something, not even when your name is Albert. So maybe neither Einstein nor Minkowski thought about a new quantity and therefore the concept of relativistic mass was unclear and problematic. And Indeed mass is considered a scalar, while 'Relativistic mass' is considered a vector. That is also a reason why relativistic mass can't be the same quantity as mass. Isn't it beautifull that we don't have to throw away this 'relativistic mass' but that in a different form it can be incorporated into the energy-momentum-mass relationship?

The mass in 'mv' we can call 'motion mass'. The same arguments against 'relativistic mass' we can use against this 'motion mass'. mass is a scalar quantity and this 'motion mass' happens to be a vector quantity. Suppose that someone introduced a new quantity instead of this 'motion mass', he calls this new quantity momentum p. Wouldn't it solve a lot of problems?

Therefore it is very likely that relativistic mass, because it isn't really mass, must be replaced by the quantity 's'.

Skip your traditional method, the equivalence of the metrics gives the hypercomplex structure:

$$\frac{E^2}{c^4} + s_x^2c^2 + s_y^2c^2 + s_z^2c^2 = m^2 + \frac{p_x^2}{c^2} + \frac{p_y^2}{c^2} + \frac{p_z^2}{c^2}$$

This equivalence of two metrics can be described by the product of two opposite quaternions and their conjugates:

$$( \frac{E}{c^2} + \mathbf{i} s_xc + \mathbf{j}s_yc + \mathbf{k}s_zc) ( \frac{E}{c^2} - \mathbf{i} s_xc - \mathbf{j}s_yc - \mathbf{k}s_zc) = ( - m - \mathbf{i} \frac{p_x}{c} - \mathbf{j} \frac{p_y}{c} - \mathbf{k} \frac{p_z}{c}) ( - m + \mathbf{i} \frac{p_x}{c} + \mathbf{j} \frac{p_y}{c} + \mathbf{k} \frac{p_z}{c})$$

Show me were I introduced self-inconsistencies in my arguments.

It is a good thing being critical, but one must also have an open mind.
Not all of physics received it's final version yet.

Dimsun

14. Feb 8, 2010

### torquil

What you have neglected to consider is that momentum is actually defined like this:

In the x-direction

$$p_x := \gamma(v)mv_x = m_{rel}v_x$$

and in the time-direction:

$$E := \gamma(v)m = m_{rel}$$

NB: m is restmass, and I use units where c=1. I never write the relativistic mass because it is not a necessary algebraic symbol. It is identical to the energy E, apart from trivial factors of c.

$$\gamma(v) := \frac{1}{\sqrt{1-v^2}}$$

Therefore, your algebra trick doesn't add anything new. What you are just saying is that

$$m_{rel}^2 - p_x^2 - p_y^2 - p_z^2 =E^2 - p^2_x - p^2_y - p^2_z$$

This is nothing new, since we already know that E=m_{rel}. Your $$s$$-variables are in essence just momenta, possibly multiplied by some power of c, depending on how you defined them. This is why I said you had written 0=0, or x=x. It is an equation that is satisfied independently of the values of the individual quantities in the equation, because you have exactly the same expression on both side. The fact that your equation is trivially satisfied is an expression of self-consistency of the different definitions in SR.

Please look up the connection between momentum end velocity in a basic exposition of SR, and you will see that you have not invented a new variable. The PDF at the top of the following page seems to be a good document, although I haven't look through it in detail:
http://galileo.phys.virginia.edu/classes/252/

Btw, relativistic mass is not a vector, it is the time-component of the momentum 4-vector. I.e. the same as energy.

Torquil

Last edited: Feb 8, 2010
15. Feb 8, 2010

### dimsun

In the same way you can write:

... Your $$m$$-variable is in essence just energy, possibly multiplied by some power of c, depending on how you defined them. This is why I said you had written 0=0, or x=x. ...

The problem is in this "possibly multiplied by some power of c".

Dimsun

16. Feb 9, 2010

### Ich

Well, mrel is in fact just energy.

That's not a problem, as we usually use dimensions where c=1.

17. Feb 9, 2010

### torquil

No, that is actually not the problem. It like saying that a calculation is incorrect because length is measured in feet instead of measured in meters. I'm simply measuring length in units of the distance light will move in one second.

I am starting to suspect that you are just having a laugh? Or you just really don't want to let go of your theory? Feel free to rewrite my entire argument and include all factors of c in the appropriate places. It doesn't make a difference.

In M. Kaku's words: If you really believe in your theory, write it up in a paper with all details, and submit it to a journal. If it really is a revolutionary work, any journal will be eager to publish your results, since it will provide them with lots of citations to increase their impact factor and therefore popularity among scientists.

Torquil

18. Feb 9, 2010

### dimsun

I am not laughing. It is pity that you say something like that. I want to discuss with logical reasoning not with personal statements. No one is forcing you to discuss this subject with me.

You didn't convince me with your reasoning. That's all.

I never heard that relativistic mass is momentum in disguise. Maybe you can convince me with articles in which it is proven that relativistic mass and momentum are the same. But I will try to find some myself.

Dimsun

I

19. Feb 9, 2010

### torquil

I don't think I said that relativistic mass is momentum in disguise. It is energy, and one can possibly say "in disguise" because of the two factors of c. But since c is constant, it is practically the same.

This time with factors of c explicitly written:

Momentum:

$$p = \gamma(v)mv$$

where

$$\gamma(v) := \frac{1}{\sqrt{1-v^2/c^2}}$$

and energy:

$$E = \gamma(v)mc^2$$

Here, m is always rest mass. This is why the energy-mass formula looks different from the standard form (additional $$\gamma(v)$$) which uses relativistic mass:

$$E=m_{rel}c^2$$

These things are discussed in all the different references I have given above. The last reference I gave seems like a particularly good one because it has a lot of explanations on the different throught-experiments, and also real experiments that support SR.

Torquil

20. Feb 9, 2010

### dimsun

You are right, you didn't say that.

At high school, I didn't understand why momentum was a different quantity then mass. Momentum was $$mv$$, so it was mass according to me. But photon's have momentum but don't have mass. So momentum is a quantity on it's own. Now I make difference between all quantities. Mass is not energy. Some particles have mass and no energy, other particles have energy but no mass. And I also can't neglect the dimensionality of quantities, even when it is velocity. For example c is not the same as c2, they have different dimensions. But that is my personal perspective.

Relativistic energy:

$$E_{rel} = \frac{m_0c^2}{\gamma}$$

$$E_{rel} = \frac{E_0}{\gamma}$$

Relativistic mass:

$$m_{rel} = \frac{m_0}{\gamma}$$

The equivalence between rest-mass and rest-energy lacks the gamma-factor, so it is different from the equivalence between relativistic energy and rest-mass:

$$E_0 = m_0c^2$$

For rest-mass:

$$m_0c^2 = \sqrt{E_{rel}^2 - p_x^2c^2 - p_y^2c^2 - p_z^2c^2 }$$

$$m_0 = \sqrt{m_{rel}^2 - s_x^2c^2 - s_y^2c^2 - s_z^2c^2 }$$

And for rest-energy:

$$E_0 = \sqrt{E_{rel}^2 - p_x^2c^2 - p_y^2c^2 - p_z^2c^2 }$$

$$\frac{E_0}{c^2} = \sqrt{m_{rel}^2 - s_x^2c^2 - s_y^2c^2 - s_z^2c^2 }$$

And therefore:

$$c^2\sqrt{m_{rel}^2 - s_x^2c^2 - s_y^2c^2 - s_z^2c^2 } = \sqrt{E_{rel}^2 - p_x^2c^2 - p_y^2c^2 - p_z^2c^2 }$$

Dimsun