# Showing Delta^3(p-q) is Not Lorentz Invariant

• A
• TroyElliott
In summary: No.If you want to work through this properly, start by proving formula (2.34), i.e.,$$\delta\Big( f(x) - f(x_0) \Big) ~=~ \frac{1}{|f'(x)|} \; \delta( x - x_0) ~=~ \frac{1}{|f'(x_0)|} \; \delta( x - x_0) ~,$$where I have given you an extra hint by showing the middle step (which is all you really need here to perform the computation at the top of p23).Btw, the ##|\dots|## should be seen
TroyElliott
From page 22 of P&S we want to show that ##\delta^{3}(\vec{p}-\vec{q})## is not Lorentz invariant. Boosting in the 3-direction gives ##p_{3}' = \gamma(p_{3}+\beta E)## and ##E' = \gamma(E+\beta p_{3})##. Using the delta function identity ##\delta(f(x)-f(x_{0})) = \frac{1}{|f'(x_{0})|}\delta(x-x_{0}),## we get $$\delta^{3}(\vec{p}-\vec{q}) = \delta^{3}(\vec{p'}-\vec{q'}) \frac{dp'_{3}}{dp_{3}}.$$

This is where I am confused, how is this step arrived at? From what I can see we have
$$\delta^{3}(\vec{p'}-\vec{q'}) = \delta(p_{1}-q_{1})\delta(p_{2}-q_{2})\delta( \gamma((p_{3}-q_{3})+\beta (E_{p}-E_{q})).$$

We can further write $$\delta( \gamma((p_{3}-q_{3})+\beta (E_{p}-E_{q}))) = \delta((p_{3}-q_{3})+\beta (E_{p}-E_{q}))/\gamma.$$

From here I don't see how to transform ##\delta((p_{3}-q_{3})+\beta (E_{p}-E_{q}))/\gamma## into the form ##\delta(p_{3}-q_{3})\frac{dp_{3}}{dp'_{3}}.##

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TroyElliott said:
[...] delta function identity ##\delta(f(x)-f(x_{0})) = \frac{1}{|f'(x_{0})|}\delta(x-x_{0}),## we get $$\delta^{3}(\vec{p}-\vec{q}) = \delta^{3}(\vec{p'}-\vec{q'}) \frac{dp'_{3}}{dp_{3}}.$$This is where I am confused, how is this step arrived at?
Think of ##p'## as a function of ##p##. I.e., substitute ##f(x) \to p'(p)## and ##f(x_0) \to p'(q)##. Then the ##\frac{dp'_{3}}{dp_{3}}## is essentially just the term ##1/|f'(x_0)|##.

HTH.

mfb and TroyElliott
strangerep said:
Think of ##p'## as a function of ##p##. I.e., substitute ##f(x) \to p'(p)## and ##f(x_0) \to p'(q)##. Then the ##\frac{dp'_{3}}{dp_{3}}## is essentially just the term ##1/|f'(x_0)|##.

HTH.
Thanks!

So taking this route we end up with $$\delta^{3}(\vec{p}-\vec{q}) = \delta^{3}(\vec{p'}-\vec{q'})\frac{E'}{E},$$

where the ##\frac{dp'_{3}}{dp_{3}}## term became ##\frac{E'}{E}.## When using the Dirac delta identity above, we should technically evaluate ##\frac{dp'_{3}}{dp_{3}}## at ##q_{3}##, right? Does this not need to be explicitly done since we are dealing with a delta function, which will force ##p_{3} ## to equal ##q_{3}## when integrated?

Thank you very much.

TroyElliott said:
When using the Dirac delta identity above, we should technically evaluate ##\frac{dp'_{3}}{dp_{3}}## at ##q_{3}##, right? Does this not need to be explicitly done since we are dealing with a delta function, which will force ##p_{3} ## to equal ##q_{3}## when integrated?
Strictly speaking, all that delta fn stuff only makes sense in the context of the theory of distributions (aka generalized functions). But after a relation involving delta fns is established using rigorous integral expressions, most people just use shorthand notation, sans integrals.

I think it's most easily to see considering momentum-space volume elements ##\mathrm{d}^3 p##. It's important to note that we deal with "on-shell particles" for the asymptotic free states, i.e., ##p^0=E_{\vec{p}}=\sqrt{\vec{p}^2+m^2}##. This means you can effectively work with functions dependent on four-momentum with the on-shell constraint understood. Then you can consider the invariant (under proper orthochronous Poincare transformation)
$$\mathrm{d}^4 p \Theta(p^0) \delta(p^2-m^2)=\frac{\mathrm{d^3} p}{2E_{\vec{p}}},$$
which implies that
$$\delta^{(3)}(\vec{p}-\vec{p}') E_{\vec{p}}$$
is a Lorentz scalar.

strangerep said:
Strictly speaking, all that delta fn stuff only makes sense in the context of the theory of distributions (aka generalized functions). But after a relation involving delta fns is established using rigorous integral expressions, most people just use shorthand notation, sans integrals.

I still do not fully understand your answer and do not see how it answers the question. SO do we need to evaluate the derivate at ##q_3##?

I still do not fully understand your answer and do not see how it answers the question. So do we need to evaluate the derivate at ##q_3##?
No.

If you want to work through this properly, start by proving formula (2.34), i.e.,
$$\delta\Big( f(x) - f(x_0) \Big) ~=~ \frac{1}{|f'(x)|} \; \delta( x - x_0) ~=~ \frac{1}{|f'(x_0)|} \; \delta( x - x_0) ~,$$where I have given you an extra hint by showing the middle step (which is all you really need here to perform the computation at the top of p23).

Btw, the ##|\dots|## should be seen as the modulus of a Jacobian -- this is necessary in general for the multivariate case.

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vanhees71

## 1. What is Delta^3(p-q)?

Delta^3(p-q) is a mathematical expression used in the study of Lorentz transformations, which are used to describe the behavior of objects in special relativity. It represents the difference between two four-vectors, p and q, in a three-dimensional space.

## 2. How is Delta^3(p-q) related to Lorentz invariance?

Delta^3(p-q) is used to show that certain physical quantities, such as energy and momentum, are not invariant under Lorentz transformations. In other words, their values appear different when measured from different frames of reference, which goes against the principle of Lorentz invariance.

## 3. What does it mean to say that Delta^3(p-q) is not Lorentz invariant?

When we say that Delta^3(p-q) is not Lorentz invariant, we mean that its value changes when we apply a Lorentz transformation to it. This indicates that the quantity it represents is not conserved or preserved in all frames of reference.

## 4. Why is it important to show that Delta^3(p-q) is not Lorentz invariant?

The principle of Lorentz invariance is a fundamental concept in special relativity, and it has been extensively tested and confirmed by experiments. By showing that Delta^3(p-q) is not Lorentz invariant, we can identify areas where our understanding of the behavior of objects in special relativity may be incomplete or incorrect.

## 5. How is the fact that Delta^3(p-q) is not Lorentz invariant used in scientific research?

The non-invariance of Delta^3(p-q) is used to develop more accurate theories and models in the field of special relativity. It also helps us to better understand the behavior of particles at high speeds and in different frames of reference, which has important implications for fields such as particle physics and cosmology.

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