# How to deduce this formula

1. Aug 20, 2007

### unica

I encountered a problem during studying QFT,can anyone help me out?

How to directly deduce this formula as follow:
$$g_{\mu\nu}\Lambda^{\mu}_{\rho}\Lambda^{\nu}_{\sigma}=g_{\rho\sigma}$$

where g is a Minkowski metric and$$\Lambda$$ is a Lorentz transformation matrix.

I am looking forward to seeing the answer which is deduced directly from the left side to the right side.Thank you

Last edited: Aug 20, 2007
2. Aug 20, 2007

### dextercioby

How does the covariant components of a 2-nd rank tensor on Minkowski spacetime behave/transform when subject to a Lorentz transformation ?

3. Aug 20, 2007

### StatusX

You can't really derive that - that is precisely the definition of a lorentz transformation.

4. Aug 20, 2007

### unica

Yes,I know it imply that the interval of two points is invariant in the Lorentz transformation.

I have a plausible deduction as follow:
Because$$g=\left(\begin{array}{cccc}-1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{array}\right),\Lambda=\left(\begin{array}{cccc}\gamma&-\beta\gamma&0&0\\-\beta\gamma&\gamma&0&0\\0&0&1&0\\0&0&0&1\end{array}\right)$$
So we can directly do the prodcut of g and two $$\Lambda$$,and this product does equal to g.

But there is an error in the last deduction.Because of the difference of the relative velocity of that two$$\Lambda$$,we cannot use the same$$\beta$$and$$\gamma$$ in that two$$\Lambda$$,and we should use $$\beta$$,$$\gamma$$ and $$\beta^{\prime}$$and$$\gamma^{\prime}$$respectively.After this revision,I cannot deduce the formula.

So I ask how to deduce it?

Last edited: Aug 20, 2007
5. Aug 20, 2007

### robphy

It might be helpful to use rapidities:
$$\beta=\tanh\theta$$, $$\gamma=\cosh\theta$$ and $$\beta\gamma=\sinh\theta$$.
It might help recognizing identities among the "relativistic factors".

If you do matrix multiplication, if I'm not mistaken, you should be evaluating $$\Lambda^\top g \Lambda$$

6. Aug 20, 2007

### unica

But even using the rapidity,we still encounter the key point,that is ,the different rapidities between that two $$\Lambda$$,you can see that the one is$$\Lambda^{\mu}_{\rho}$$,the other is$$\Lambda^{\nu}_{\sigm a}$$,so during deduction we should use $$\theta$$ and $$\theta^{\prime}$$respectively,and this revision will make us fail in reaching the right result

7. Aug 20, 2007

### George Jones

Staff Emeritus
You don't. As StatusX said, this is the *definition* of a Lorentz transformation, and definitions aren't deduced.

You might want to verify that Lorentz transformations seen in elementary treatments of special relativity satisfy this equation. Then, the two $\Lambda$'s are not representation of two different Lorentz transformation, they are representations of the same Lorentz transformation.

Also, note that the Lorentz transformations seen in elementary treatments of special relativity are example of special Lorentz transformations called boosts. A general (proper, orthochronous) Lorentz transformation $\Lambda$ can alway be written as the product of a boost $B$ with a rotation $R$,

$$\Lambda = BR$$.

8. Aug 20, 2007

### George Jones

Staff Emeritus
This might help, or it might confuse the issue further.

Let $\mathbf{x}$ be a 4-vector and $\Lambda$ be a Lorentz transformation. If $\mathbf{x}' = \Lambda \mathbf{x}$, then $\mathbf{x}' \cdot \mathbf{x}' = \mathbf{x} \cdot \mathbf{x}$, i.e.,

$$\left( \Lambda \mathbf{x} \right) \cdot \left( \Lambda \mathbf{x} \right ) = \mathbf{x} \cdot \mathbf{x}.$$

Note that both $\Lambda$'s are the same since each $\mathbf{x}'$ is the same. It is not the case that one $\Lambda$ is transformed with respect to the other $\Lambda$.

The equation that you wrote down is a component verstion of the above equation with the arbitrary $\mathbf{x}$'s "divided out", and with the dots replaced by g's.

Last edited: Aug 20, 2007
9. Aug 20, 2007

### unica

Thank you for Jones's response.But I am still confused.

Yes,I want to verify that Lorentz transformation seen in elementary treatments of special relativity satisfy this equation.And I accept that those two $\Lambda$s represent the same Lorentz transformation.

But the $$\Lambda^{\mu}_{\rho}$$ originates from$$\overline{x}^{\mu}=\Lambda^{\mu}_{\rho}x_{\rho}$$,
the$$\Lambda^{\nu}_{\sigma}$$ originates from$$\overline{x}^{\nu}=\Lambda^{\nu}_{\sigma}x_{\sigma}$$,
and we can obviously see that the two relative velocities in the two $$\Lambda$$s are generally not the same.So why these two$$\Lambda$$ represent the same Lorentz transformation?I admit that they are identical in frame but different in parameter.

so we can write them as follow:
$$\Lambda^{\mu}_{\rho}=\left(\begin{array}{cccc}\gamma&-\beta\gamma&0&0\\-\beta\gamma&\gamma&0&0\\0&0&1&0\\0&0&0&1\end{array }\right)$$
and
$$\Lambda^{\nu}_{\sigma}=\left(\begin{array}{cccc}\gamma^{\prime}&-\beta^{\prime}\gamma^{\prime}&0&0\\-\beta^{\prime}\gamma^{\prime}&\gamma^{\prime}&0&0\\0&0&1&0\\0&0&0&1\end{array }\right)$$
and if you do the product as follow:
$$\left(\begin{array}{cccc}\gamma&-\beta\gamma&0&0\\-\beta\gamma&\gamma&0&0\\0&0&1&0\\0&0&0&1\end{array }\right)\left(\begin{array}{cccc}-1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{array}\right)\left(\begin{array}{cccc}\gamma^{\prime}&-\beta^{\prime}\gamma^{\prime}&0&0\\-\beta^{\prime}\gamma^{\prime}&\gamma^{\prime}&0&0\\0&0&1&0\\0&0&0&1\end{array }\right)$$
you cannot achieve the g of right hand side.But if we delete all the primes ,everything will go right.

Then,I am lack of the knowledge of the general Lorentz transformation.Can you give me some hints or links?Thank you very much.

Last edited: Aug 20, 2007
10. Aug 20, 2007

### unica

I have understood your mean.Thank you very much,George.