Show d^4k is Lorentz Invariant

Show that

d^4k

is Lorentz Invariant

2. Relevant equations

Under a lorentz transformation the vector

k^u

transforms as

k'^u=\Lambda^u_v k^v

where

\Lambda^u_v

satisfies

\eta_{uv}\Lambda^{u}_{p}\Lambda^v_{o}=\eta_{po}

\eta_{uv}

(2) the Minkowski metric, invariant.

3. The attempt at a solution

I think my main issue lies in what

d^4k

is and writing this in terms of

d^4k

Once I am able to write

d^4k

in index notation I might be ok.

For example to show

ds^2=dx^udx_u

is invariant is pretty simple given the above identities and my initial step would be to write it as

ds^2=\eta_{uv}dx^udx^v

in order to make use (2).

I believe

d^4k=dk_1 dk_2 dk_3 dk_4

?

For example, more generally, given a vector

V^u = (V^0,V^1,V^2,V^3)

I don't know how I would express

V^0V^1V^2V^3

as some sort of index expression of

V^u

(and probably I'm guessing the Minkowski metric?).

I would like to do this for

d^4k

, if this is the first step required ?and how do I go about it? Many thanks in advance.

(No time dilation, length contraction argument please, I need to make use of what is given in the question - thank you).

Reply 1

atsruser

Original post by xfootiecrazeesarax

Show that

d^4k

is Lorentz Invariant

I'm not sure what you mean by

d^4k

- is this working in momentum space or something?

Anyway, I've seen this done for the usual 4-volume element by Jacobians. You have under a Lorentz transformation:

d^4 x'= dx' dy' dz' dt' = |\mathcal{L}| dx dy dz dt

where

|\mathcal{L}|

is the determinant of the Jacobian of the Lorentz matrix. So if you can show that this is 1, then the volume elements are the same in both coord systems and I don't think that is too tricky, though I'm not sure I've ever done it myself.

Reply 2

xfootiecrazeesarax

Original post by atsruser

I'm not sure what you mean by

d^4k

- is this working in momentum space or something?

Anyway, I've seen this done for the usual 4-volume element by Jacobians. You have under a Lorentz transformation:

d^4 x'= dx' dy' dz' dt' = |\mathcal{L}| dx dy dz dt

where

|\mathcal{L}|

I want to make use of these though

Under a lorentz transformation the vector

k^u

transforms as

k'^u=\Lambda^u_v k^v

where

\Lambda^u_v

satisfies

\eta_{uv}\Lambda^{u}_{p}\Lambda^v_{o}=\eta_{po}

,

I believe they do the same as the Jacobian so I will eventually reduce to that stage probably , I suspect via an expression including the Minkowski metrics, perhaps a delta arising, however so I am still stuck on an initial index expression for

dk^1dk^2dk^3dk^4

to get me started.

1+1=3

thread broken

Original post by xfootiecrazeesarax

I want to make use of these though

Under a lorentz transformation the vector

k^u

transforms as

k'^u=\Lambda^u_v k^v

What is

k^u

? The relativistic wave vector? i.e. are you working with

\bold{k} = (\frac{\omega}{c}, k_x, k_y, k_z)

?

In that case, then I guess that

d^4k = \frac{d\omega}{c} dk_x dk_y dk_z

I also have no idea what your

\eta

symbols are. Can you please explain this stuff clearly. You are leaving too many details out. I can't read your mind and I have no idea what you have been working on.

Reply 5

EternalLight

Original post by xfootiecrazeesarax

I want to make use of these though

Under a lorentz transformation the vector

k^u

transforms as

k'^u=\Lambda^u_v k^v

where

\Lambda^u_v

satisfies

\eta_{uv}\Lambda^{u}_{p}\Lambda^v_{o}=\eta_{po}

dk^1dk^2dk^3dk^4

to get me started.

I suggest you try posting on https://www.physicsforums.com/ You may get a lot more replies.

Reply 6

xfootiecrazeesarax

Original post by EternalLight

I suggest you try posting on https://www.physicsforums.com/ You may get a lot more replies.

done haha
and no such luck :wink:

Reply 7

Farhan.Hanif93

Original post by xfootiecrazeesarax

I want to make use of these though

Under a lorentz transformation the vector

k^u

transforms as

k'^u=\Lambda^u_v k^v

where

\Lambda^u_v

satisfies

\eta_{uv}\Lambda^{u}_{p}\Lambda^v_{o}=\eta_{po}

dk^1dk^2dk^3dk^4

to get me started.

For each

i

, define the 4-vectors

dK^{\mu}_i = \delta^{\mu}_i dk^i

(where, in case it isn't obvious, the summation convention does not apply over

i

). Then note:

d^4k = dk^0 dk^1 dk^2 dk^3 = \epsilon_{\mu \nu \sigma \rho} dK^{\mu}_0 dK^{\nu}_1 dK^{\sigma}_2 dK^{\rho}_3

Which is clearly invariant, as the object on the RHS is obviously a scalar (given that there are no free indices). If you really want to prove it using the language of tensors, you need only know how the 4-dimensional Levi-Civita symbol transforms.

That all said, as atsruser points out, the standard and most direct way to prove invariance here is to show that the Jacobian of the transformation is 1.

(edited 6 years ago)

Reply 8

Farhan.Hanif93

Original post by atsruser

What is

k^u

? The relativistic wave vector? i.e. are you working with

\bold{k} = (\frac{\omega}{c}, k_x, k_y, k_z)

?

In that case, then I guess that

d^4k = \frac{d\omega}{c} dk_x dk_y dk_z

It is invariant for all (reasonably defined) 4-vectors

k^{\mu}

I also have no idea what your

\eta

symbols are. Can you please explain this stuff clearly. You are leaving too many details out. I can't read your mind and I have no idea what you have been working on.

I'd have though

\eta

is fairly standard notation for Minkowski metric, which defines the class of Lorentz transformations. As a 4x4 matrix,

\eta_{\mu \nu} = \text{diag} (1, -1, -1, -1)_{\mu \nu}

.

Wouldn't have hurt if the OP pointed this out, however!

Reply 9

atsruser

Original post by Farhan.Hanif93

It is invariant for all (reasonably defined) 4-vectors

k^{\mu}

.

Yes. However, the OP seemed to be asking for an explicit form for

d^4k

, and I just wanted to be sure what she was dealing with.

And I'm intrigued by your "reasonably defined" caveat - are there some 4-vectors where the argument doesn't work? I can't see how.

I'd have though

\eta

is fairly standard notation for Minkowski metric, which defines the class of Lorentz transformations. As a 4x4 matrix,

\eta_{\mu \nu} = \text{diag} (1, -1, -1, -1)_{\mu \nu}

I had forgotten this, and had to look it up, but yes, you're right.

Given the information that the OP supplied in her first post, I was wondering if they were expecting some kind of explicit matrix manipulation argument based on the eta identity. I've forgotten too much of this stuff to see if it can be done that way, though.

Reply 10

Farhan.Hanif93

Original post by atsruser

Yes. However, the OP seemed to be asking for an explicit form for

d^4k

, and I just wanted to be sure what she was dealing with.

Perfectly fair enough.

And I'm intrigued by your "reasonably defined" caveat - are there some 4-vectors where the argument doesn't work? I can't see how.

It was a little unclear - by that, I meant 4-vectors

k^{\mu}

for which we can define the differential element

dk^{\mu}

as something sensible. It was more a comment to cover my back in case I had missed a case. For almost all 4-vectors, it should be fine.

Given the information that the OP supplied in her first post, I was wondering if they were expecting some kind of explicit matrix manipulation argument based on the eta identity. I've forgotten too much of this stuff to see if it can be done that way, though.

Yes, that is indeed what they seem to be after. (Although, it would help if the OP verified this).

In particular, from my post above, OP wants to show that:

\epsilon'_{\mu \nu \sigma \rho} dK'^{\mu}_0 dK'^{\nu}_1 dK'^{\sigma}_2 dK'^{\rho}_3 = \epsilon_{\mu \nu \sigma \rho} dK^{\mu}_0 dK^{\nu}_1 dK^{\sigma}_2 dK^{\rho}_3

where:

dK'^{\mu}_i = \Lambda^{\mu}_{\nu}dK^{\nu}

\epsilon'_{\mu \nu \sigma \rho} = \Lambda_{\mu}^{\alpha}\Lambda_{ \nu}^{\beta} \Lambda_{\sigma}^{\gamma} \Lambda_{\rho}^{\theta}\epsilon_{\alpha \beta \gamma \theta}

A slightly tedious argument which explicitly requires use of the relationship between the Minkowski metric and Lorentz transformations.

(edited 6 years ago)

Reply 11

atsruser

Original post by Farhan.Hanif93

It was a little unclear - by that, I meant 4-vectors

k^{\mu}

for which we can define the differential element

dk^{\mu}

as something sensible. It was more a comment to cover my back in case I had missed a case. For almost all 4-vectors, it should be fine.
By sensible, you mean physically meaningful, or useful? I guess there must be 4-vectors where the concept of an associated 4-volume isn't very productive.

As for the rest, your mad index skilz are over my head at the moment. I'm too rusty to decode it.