/sci/ - Science & Math

[Anonymous No.3575437]

What's the difference between using

E = mc^2
and
E^2 = (mc^2)^2 + (pc)^2?

[Anonymous No.3575445]

Image displays a bad understanding of the uncertainty principle

This is unacceptable

[Anonymous No.3575446]

>>3575437
In the first equation m is the relativistic mass and in the second equation m is the rest mass.

[Anonymous No.3575464]

>>3575445

Seriously.

It should be "Schroedinger".

[Anonymous No.3575475]

>>3575446
is the momentum relativistic?

[Anonymous No.3575482]

The 1st one is only true for particles who have 0 momentum.

[Anonymous No.3575495]

>>3575445
you realize QM has so many different philosophical interpretations, that perhaps OP's image is valid... dont be such a fool... whether or not the writer knew what they were saying will never be possible to know.

[Anonymous No.3575505]

>>3575437
the first only count when p =0 , the particle is at rest. the other one always counts.

[Anonymous No.3575509]

>>3575464
Then it would be demonstrating a poor understanding of superposition.

[Anonymous No.3575531]

The first one is mass-energy equivalence, and doesn't always have special meaning. No, photons do not have mass - it just tells you how much mass is equivalent to a given amount of energy.

The second one is the relationship between total energy, rest mass, and momentum.

[Anonymous No.3575590]

>>3575531
>>3575505
>>3575482
>>3575446
can you equate them at all taking into account the lorentz factor?

[Anonymous No.3575604]

>>3575590
yes, the energy is different in different frames. its quite easy to do.

[Anonymous No.3575626]

Unless the particle is moving at relativistic speeds, pc will always be very small compared to mc^2, so you can ignore its effects in most cases.

[Anonymous No.3575634]

hint: what condition of the second makes the first one?

[Anonymous No.3575641]

im just going to leave this link here

http://galileo.phys.virginia.edu/classes/252/energy_p_reln.html

[Anonymous No.3575676]

the second equation is the general equation.

The mass term is rest mass, the momentum term has the gamma factor built in because it contains a velocity factor.

Thus, when the velocity is C, the mass term goes away and you are left with only the momentum equation, E = pc, i.e. the case of the photon

When the velocity goes to 0, you are left with the rest mass equation E = mc^2.

[Anonymous No.3575712]

>>3575437
E^2 = (mc^2)^2 + (pc)^2 is the general correct expression.

E^2 = (mc^2)^2 is only valid is p=0

anything else?

[Anonymous No.3575720]

can someone tell me whether this is correct

[Anonymous No.3575740]

can someone tell me whether this is correct

sorry for not using latex, i really need to learn it

[Anonymous No.3575763]

>>3575740
>>3575740

No, it is not correct.

Maybe show us the problem and we can help you solveit.

[Anonymous No.3575776]

>>3575740
>>3575740
p = mv is only true in classical mechanics, not relativity. it should be p = (gamma)mv

[Anonymous No.3575792]

>>3575776
no but i thought the second equation used rest mass, therefore the second equation is
E = (mc^2)^2 + (mvc)^2

and the first one is
E = (gamma)mc^2

i may be totally wrong...

[Anonymous No.3575803]

>>3575792
if you are using rest mass and relativistic mass then you cant cancel it out in line 6.

[Anonymous No.3575818]

>>3575803
i'm not using different masses, i'm using rest mass throughout
in the first equation i've split relativistic mass into gamma and rest mass

[Anonymous No.3575822]

E=mc^2...
BUT ONLY IN THE REST FRAME.

[Anonymous No.3575855]

<div class="math">m_{rel}=\gamma m_0</div><div class="math">E=m_{rel} c^2</div><div class="math">E^2=(m_0 c^2)^2+(pc)^2</div><div class="math">p=m_{rel} v=\gamma m_0 v</div>

Is that so hard to understand?