On Four Velocity and Four Momentum

[Anamitra Palit]

Certain issues regarding four velocity and four momentum have been discussed in the attached file. Requesting the audience to consider and address these issues..

Four Velocity 102.pdf

[joigus]

Can you summarize it here, please?

[Bufofrog]

3 hours ago, Anamitra Palit said:

Certain issues regarding four velocity and four momentum have been discussed in the attached file. Requesting the audience to consider and address these issues.

It is against the forum rules to just post a link and not present your idea in your post.

Edited December 18, 2020 by Bufofrog

[Phi for All]

!

Moderator Note

Please summarize here. Members shouldn't be required to click any links in order to participate.

 

[Anamitra Palit]

Thanks for your advice/instruction

In this article we have proved:

1)   Four dot product of velocities v1.v2>=c^2 [equation (2) has been derived]

2)   Four dot product of momenta p1.p2>=m1 m2 c^4 [Equation (6)]

3)   Finally we have brought out a result:v.v=c^2[gamma^2-1+1/gamma^2] [eq (7)) which contradicts standard theory

Some typos and remediable errors [of a minor nature though] have been attended  to  and corrected. The paper has been reposted below

https://drive.google.com/file/d/1-fH2BcyO5QVljei90fq49mfNm0oKzYKp/view?usp=sharing

Four Four.pdf

Edited December 18, 2020 by Anamitra Palit

[Ghideon]

Question: do you assume |a1| >=| a2| and |b1| >= |b2|?

From your attached paper: 

 

Without |a1| >=| a2| and |b1| >= |b2| the square roots may be imaginary. Is that intended?

From your attached paper:

 

[Anamitra Palit]

Looking at the application there is no problem

We have considered the signature(+,-,-,-)

Indeed v.v=c^2 (v_t)^2- (v_x)^2-(v_y)^2-(v_z)^2=(c^2)v_t^2-|v|^2=c^2

Therefore (c^2)v_t^2-|v|^2]>0

The quantities under the square root sign ,as pointed out , will be positive.That aligns itself with the application.

Thanking Ghideon for his comment.

 

Edited December 18, 2020 by Anamitra Palit

[Ghideon]

21 minutes ago, Anamitra Palit said:

The quantities under the square root sign ,as pointed out , will be positive.That aligns itself with the application.

Thanks, I'll check that later. Here is another one:

Is that correct? It seems to imply that if 2>1 then 2<-1 which is not true. 

[Anamitra Palit]

What Ghideon has interpreted is not correct.

x^2>=a^2 implies x>=|a| or x=<-|a|.

In a similar vein we may write

(a1 b1-a2 b2)^2>=(a1^2-a2^2)(b1^2-b2^2)

implies

a1 b1-a2 b2>=Sqrt[(a1^2-a2^2)(b1^2-b2^2] or a1 b1-a2 b2<=-Sqrt[(a1^2-a2^2)(b1^2-b2^2)

One has to be careful about the 'or' connective.

 

Edited December 18, 2020 by Anamitra Palit

[Ghideon]

1 hour ago, Anamitra Palit said:

One has to be careful about the 'or' connective.

True, the statement seems ok (roots of quadratic equation)

[Markus Hanke]

14 hours ago, Anamitra Palit said:

1)   Four dot product of velocities v1.v2>=c^2

This is meaningless. If you are projecting a shorter 4-vector onto a longer one, the result can never exceed the length of the longer vector, that’s just basic geometry. In other words, the longest a projection can ever be is that of a 4-vector onto itself, which, in Minkowski spacetime, is thus -1. Since the inner product is invariant, this is true for all 4-vectors in all frames.

Consider a general 4-velocity of the type

\[u^{\mu } =( \gamma ,\gamma v_{x} ,\gamma v_{y} ,\gamma v_{z})\]

The inner product with itself is

\[u^{\mu } u_{\mu } =-\gamma ^{2}\left( 1-v^{2}\right) =-\gamma ^{2} /\gamma ^{2} =-1\]

as expected. So your claim is wrong.

14 hours ago, Anamitra Palit said:

2)   Four dot product of momenta p1.p2>=m1 m2 c^4 [Equation (6)]

If this were true, the momentum of a photon would be zero. This is evidently false, as we know already from experiment and observation. Mathematically, you can show this in a similar manner as above.
So again, you are wrong.

14 hours ago, Anamitra Palit said:

3)   Finally we have brought out a result:v.v=c^2[gamma^2-1+1/gamma^2]

I’ve already shown above that the inner product of a 4-velocity with itself is -1, so the magnitude of a 4-velocity in spacetime is always exactly c. Since the inner product is invariant, this is true in all frames, so it can’t be a function of the gamma factor. Also, if you look at this expression, you should notice immediately that the resultant magnitude of v does not correspond to the gamma factor you are inputting, so the expression is meaningless.
You are wrong on this one, too.

14 hours ago, Anamitra Palit said:

which contradicts standard theory

Yes, because all three points you have presented contradict both basic maths, as well as observation in the real world. It’s simply wrong.

Edited December 19, 2020 by Markus Hanke

[joigus]

I haven't had time to review in detail. The numeric relation seems correct to me.

As to the claimed contradiction, I haven't got around to it yet.

Contradiction with what exactly?

Don't have time to read Markus' comments. Maybe later.

[joigus]

Physical four velocities are not made of arbitrary numbers. As Markus said, in just a bit more explicit notation and rephrasing what he said, although I think it was clear enough,

\[ \left( v_t, v_x, v_y, v_z \right ) \]

but constrained to,

\[ \left( \frac{c}{\sqrt{1-v^2/c^2}},  \frac{v_x}{\sqrt{1-v^2/c^2}}, \frac{v_y}{\sqrt{1-v^2/c^2}},  \frac{v_z}{\sqrt{1-v^2/c^2}} \right) \]

Same with 4-momenta. 4-momentum is the product of \( u^{\mu} \) --the 4-velocity-- times the mass. So it doesn't do to fill arbitrary numbers in the slots, so to speak.

Edited December 19, 2020 by joigus

[Anamitra Palit]

9 hours ago, Markus Hanke said:

This is meaningless. If you are projecting a shorter 4-vector onto a longer one, the result can never exceed the length of the longer vector, that’s just basic geometry. In other words, the longest a projection can ever be is that of a 4-vector onto itself, which, in Minkowski spacetime, is thus -1. Since the inner product is invariant, this is true for all 4-vectors in all frames.

Consider a general 4-velocity of the type

 

uμ=(γ,γvx,γvy,γvz)

 

The inner product with itself is

 

uμuμ=−γ2(1−v2)=−γ2/γ2=−1

 

as expected. So your claim is wrong.

If this were true, the momentum of a photon would be zero. This is evidently false, as we know already from experiment and observation. Mathematically, you can show this in a similar manner as above.
So again, you are wrong.

I’ve already shown above that the inner product of a 4-velocity with itself is -1, so the magnitude of a 4-velocity in spacetime is always exactly c. Since the inner product is invariant, this is true in all frames, so it can’t be a function of the gamma factor. Also, if you look at this expression, you should notice immediately that the resultant magnitude of v does not correspond to the gamma factor you are inputting, so the expression is meaningless.
You are wrong on this one, too.

Yes, because all three points you have presented contradict both basic maths, as well as observation in the real world. It’s simply wrong.

Markus Hanke has applied classical commonsense to interpret pseudo Riemannian geometry. when he talks of the projection issue

Markus Hanke  claims

"This is meaningless. If you are projecting a shorter 4-vector onto a longer one, the result can never exceed the length of the longer vector, that’s just basic geometry."

Classically if the norm of a vector is zero its components are individually zero. But this is not true of the null vector in pseudo Riemannian geometry.. Its norm is zero but the components are non zero. The null vector for the pseudo Riemann case,is parallel to itself and perpendicular to itself simultaneously[remember , the components are in zero in general].Can you prove mathematically  the claim staked by Markus v1.v2<c^2? On the contrary the paper has derived mathematically v1.v2>=c^2;  v1=v2 corresponds to the case v.v=c^2 have

Next: It has been proved mathematically in the paper:

Four dot product of momenta p1.p2>=m1 m2 c^4

Pl note (E/c)^2-|p|^2=m^2c^2 [implies E^2-|p|^2 c^2=m^2 c^4].

We do have from the above  that is from (E/c)^2-|p|^2=m^2c^2, p.p=m^2c^2

In the paper we have derived

E1E2-c^2|p1||p2|>=m1m2c^4

That again leads us to v.v=c^2[gamma^2-1+1/gamma^2]

 

Formal consideration of photon momentum [as in standard theory]

Now p.p=E^2-c^2|p|^2=m^2 c^4

Setting rest mass= zero on the right side only we obtain:E=pc

[NB:On the left side of the last equation,with the rest mass tending to zero and gamma tending to infinity for the photon , the product m=m_0 gamma matches with the finite value on the right side] 

 But from quantum mechanics: E=h nu=hc/lambda or pc=hc/lambda or p=h/lambda not equal to zero.

[nu=frequency,h:Planck's constant, lambda=wavelength]

For my derivation Markus thinks momentum of the photon = zero why?

It is important to keep in the mind that the results derived in the paper have finally led to a discrepancy:

v.v=c^2[gamma^2-1+1/gamma^2]

It stands in contradiction to conventional  results which require v.v=c^2. The very intention of the paper is to highlight this contradiction in the theory.

Edited December 19, 2020 by Anamitra Palit

[Anamitra Palit]

The paper has been reposted with some minor corrections:basic conclusions remain unchanged

We have p1.p2>=m1m2 c^2 

The relation

v.v=c^2[gamma^2-1+1/gamma^2] remains unchanged

 

Four Four 3.pdf

Edited December 19, 2020 by Anamitra Palit

[Markus Hanke]

13 hours ago, Anamitra Palit said:

Can you prove mathematically  the claim staked by Markus v1.v2<c^2?

This is not what I said - you need to go back and look at what I actually said, and you will also find the maths there.

13 hours ago, Anamitra Palit said:

For my derivation Markus thinks momentum of the photon = zero why?

Photons have no rest mass, so according to your expression, the inner product of photon momentum with itself is zero. Since the inner product can vanish only if the two vectors are either perpendicular, or one of the vectors has zero magnitude, that means that according to you the photon has no net momentum.

12 hours ago, Anamitra Palit said:

The paper has been reposted with some minor corrections

Just repeating the same thing again does not make it any less wrong.  

[Anamitra Palit]

Makus Henke has remarked;

"Photons have no rest mass, so according to your expression, the inner product of photon momentum with itself is zero. Since the inner product can vanish only if the two vectors are either perpendicular, or one of the vectors has zero magnitude, that means that according to you the photon has no net momentum."

 

Four momentum of a photon[real photon] is zero. This is the mass-shell condition for the real photon.But the spatial part of the photon momentum is not zero

Indeed,

E^2-c^2|p|^2=m_0^2c^2[m_0 is the rest mass]

(m_0 gamma)^2 c^2 -(m_0 gamma()^2=m_0^2c^2

The rest mass m_0  for the photon is zero . WE cannot cancel zero from the two sides. We simply write 

(m_0 gamma)^2 c^2 -(m_0 gamma()^2=0

For the photon m_0 gamma is of the form zero * infinity. We have

E^2=c^2|p|^2 or E=|p|c

From quantum mechanics E=hc/lambda

|p|c=hc/lambda

or,|p|=h/lambda,a non zero value

Markus Henke has remarked

"Just repeating the same thing again does not make it any less wrong. "

If he finds anything wrong with my paper  he should point to it in a specific manner. So far I have refuted everything has has considered incorrect with my paper

 

 

 

[Anamitra Palit]

Makus Henke has remarked;

"Photons have no rest mass, so according to your expression, the inner product of photon momentum with itself is zero. Since the inner product can vanish only if the two vectors are either perpendicular, or one of the vectors has zero magnitude, that means that according to you the photon has no net momentum."

 

Norm of the four momentum of a photon[real photon] is indeed zero. This is the mass-shell condition for the real photon.But for the spatial part of the photon ,norm  of momentum is not zero[this portion in italics is a revision over the last post]

Indeed,

E^2-c^2|p|^2=m_0^2c^2[m_0 is the rest mass]

(m_0 gamma)^2 c^2 -(m_0 gamma()^2=m_0^2c^2

The rest mass m_0  for the photon is zero . WE cannot cancel zero from the two sides. We simply write 

(m_0 gamma)^2 c^2 -(m_0 gamma)^2=0

For the photon m_0 gamma is of the form zero * infinity.Considering it to be a non zero finite quantity [this italized prt bis an addition to the earlier post]we have

E^2=c^2|p|^2 or E=|p|c

From quantum mechanics E=hc/lambda

|p|c=hc/lambda

or,|p|=h/lambda,a non zero value

Markus Henke has remarked

"Just repeating the same thing again does not make it any less wrong. "

If he finds anything wrong with my paper  he should point to it in a specific manner. So far I have refuted everything has has considered incorrect with my paper

 

• Quote

Edited December 20, 2020 by Anamitra Palit

[Markus Hanke]

22 hours ago, Anamitra Palit said:

So far I have refuted everything has has considered incorrect with my paper

You haven’t. You are simply repeating the same statements again and again.

22 hours ago, Anamitra Palit said:

Norm of the four momentum of a photon[real photon] is indeed zero.

You are right, this was my mistake, I actually thought of the photon’s 3-momentum (which is not zero!) when I wrote my post, whereas you were referring to the 4-momentum.
 

 

[Anamitra Palit]

Markus Hanke's Points have been clearly refuted. He could not find errors with my calculations. Something has to be wrong with the conventional theory:

Some General Types of Difficulties with the theory of Tensors:

The transformation of a rank two covariant tensor has been considered. Then we proceed to consider a diagonal tensor[off diagonal components are zero:A^(mu nu)=0 for mu not equal to nu] to bring out a result that all tensors should be null tensors. A link to the google drive file has been provided.A file has also been attached considering the file attachment facility that has been provided by the forum. 

https://drive.google.com/file/d/10z63Xidgs3m8p04_C6ZiGh-8Q6KTwPsh/view?usp=sharing

Incidentally I tried the Latex with the code button.But I am not getting the correct preview.

Example

\begin{equation}\bar{A}^{\mu\nu}=\frac{\partial \bar{x}^{\mu}}{\partial x^{\alpha}}\frac {\partial \bar{x}^{\nu}}{\partial x^{\beta}}A^{\alpha \beta}\end {equation}

 

\[{\ bar{A}}^{\mu\nu}=\frac{\partial {\bar{x}}^{\mu}}{\partial x^{\alpha}}\ frac{\partial {\bar{x}}^{\nu}}{\partial x^{\beta}}\]

Currently for a long time I am not a frequent user of Latex thanks to the equation bar of MS Word. But I do appreciate,like many others, the application of Latex in various forums.Help is being requested from the forum regarding Latex.

Edited December 22, 2020 by Anamitra Palit

[Ghideon]

Two relations taken from your paper: 

Can you show that they are valid in the physics under discussion? As Joigus said:

On 12/19/2020 at 5:03 PM, joigus said:

Physical four velocities are not made of arbitrary numbers.

 

 

 

[joigus]

5 hours ago, Anamitra Palit said:

Markus Hanke's Points have been clearly refuted.

Funny that you think that, as nobody else does.

Without going into details, it's quite clear for anybody who's worked with tensor calculus for some time that the most likely thing going on is that you're confusing invariant properties with coordinate-dependent properties, and mixing them all up in one big mess.

Some tensor identities can be proved by appealing to some tensor being zero in one particular coordinate system. Then it must be zero in all coordinate systems. Conversely, non-zero in one system <=> Non-zero in all. On the contrary, the Christoffel symbols can always be chosen to be zero in one coordinate system, but non-zero in infinitely many other coordinate systems. Playing with these two properties facilitates some proofs, but you must know what you're doing. Handling index expressions without any care of what is a tensor and what only holds in one particular system is the wrong way to go.

It is a real pain to go over every step of a calculation that somebody only too obviously did wrong, because you have proven the theorems forwards and backwards and gone through all the examples. 

A tensor being diagonal, e.g., is not an invariant property under SO(3) or O(3). On the other hand, tensors like the identity \( {\left. \delta^{\mu} \right. }_{\nu} \) or \( \epsilon_{\alpha\beta\cdots} \) are called isotropic tensors, because they look the same (have the same components) in all coordinates systems. So you cannot safely assume that a diagonal tensor takes part in any tensor equation.

"Diagonal matrix" makes sense, meaning "a matrix that looks diagonal in a particular base". "Diagonal tensor" does not.

Edited December 22, 2020 by joigus

[Anamitra Palit]

2 hours ago, Ghideon said:

Two relations taken from your paper: 

Can you show that they are valid in the physics under discussion? As Joigus said:

 

 

 

The mathematical formulas expressed by (3) have been applied to obtain the  two formulas referred to by Ghideon. The application technique has been shown just after the results. And physics ,of course, is expected to cater to mathematics. One of the options[inequations] has to be valid.

Link to google drive:

https://drive.google.com/file/d/1a5Sd5Vfhbg0gqERIFDcjxXmIjZXw6ulr/view?usp=sharing

[joigus]

In fact, a Kronecker delta that is twice covariant or twice contravariant is not an isotropic tensor either. It must be 1-covariant 1-contravariant. Also arbitrary tensor products of 1-covariant 1-contravariant Kronecker deltas is an isotropic tensor. Arbitrary products are not. That's because contravariant indices transform with the inverse matrix with respect to covariant ones (that's why they're called "contra"). If you multiply twice by the same matrix you don't get back to Kronecker deltas.

You must go carefully through all these checks in order not to make elementary mistakes. It's a natural rite of passage.

The literature is full of mistakes of this nature. Not in the really prestigious books, of course.

6 minutes ago, Anamitra Palit said:

And physics ,of course, is expected to cater to mathematics

No. Mathematical physics is expected to cater to physics. Mathematics doesn't need any imput from physics.

Mathematical physics is expected to be self-consistent, and further, it is expected to comply with what we measure in the laboratory.

[Ghideon]

 

1 hour ago, Anamitra Palit said:

The mathematical formulas expressed by (3) have been applied to obtain the  two formulas referred to by Ghideon. The application technique has been shown just after the results. And physics ,of course, is expected to cater to mathematics. One of the options[inequations] has to be valid.

Please provide the details here on the forum. Please include or describe the physics that justifies the relations; the relations are not physically valid for arbitrary numbers. 

 

 

Edited December 22, 2020 by Ghideon