Deducing Formula from Lorentz Transformation Matrix

[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

 

[dextercioby]

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

 

[StatusX]

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

 

[unica]

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

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?

 

[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 $$\Lambda^\top g \Lambda$$

 

[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

 

[George Jones]

So I ask how to deduce it?

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 $R$,

$$\Lambda = BR$$.

 

[George Jones]

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.

 

[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.

 

[unica]

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