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I'm a PhD student at Queen Mary, University of London. I'm interested in modern approaches to scattering amplitudes, in particular twistor methods and integrability.


Aug
28
comment Commutator algebra in exponents
Very nicely done! I hadn't thought of doing that!
Aug
28
answered Commutator algebra in exponents
Aug
27
answered Is quantum indeterministic?
Aug
27
comment What do you exactly mean when you say that momentum is conserved?
@bobie - I think what you're really confused about is the meaning of "elastic collision". I usually take this to mean that kinetic energy is conserved, so no energy is transferred into other forms. If we had a freely moving wall and a perfect elastic collision with the ball, then kinetic energy would be transferred from the ball to the wall. This would give the wall a small velocity, enough to conserve momentum. If the wall is fixed then it cannot gain overall kinetic energy. In this case the collision is not elastic, and momentum is conserved due to (e.g.) shockwaves within the wall.
Aug
26
comment What do you exactly mean when you say that momentum is conserved?
Originally you only looked at the ball itself. If you examine the whole system you can still get an elastic collision where the wall moves very slightly. Of course in reality the collision is unlikely to be completely elastic (in the sense that kinetic energy will not be conserved in the ball-wall system, you'll also have air resistance etc. to consider).
Aug
26
comment 1-particle momentum eigenfunction in terms of field operator for real Klein-Gordon field
Could you make your question a little more precise? I don't quite understand what you're confused about. You can plug in the mode expansion for $\phi(x)$ and derive the result. Do you want to know why we should expect this result?
Aug
26
answered What do you exactly mean when you say that momentum is conserved?
Aug
19
comment Clarification of Tensor, Jacobian
Glad to be of help! If this answer clear everything up then you should accept it by clicking the green tickmark on the left hand side. This helps keep the site clean!
Aug
18
answered Clarification of Tensor, Jacobian
Aug
1
revised Stability of D-Branes and Coupling to Fields
added extra research
Aug
1
asked Stability of D-Branes and Coupling to Fields
Jul
25
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18
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Jun
25
awarded  Popular Question
Jun
18
awarded  Nice Question
May
17
comment Why isn't it $E \approx 27.642 \times mc^2$?
@ChrisWhite - you are correct that the underlying reason for $E=mc^2$ is just the postulates of special relativity. I've detailed a couple of less terse references in my answer. But I wouldn't say this has nothing to do with naturalness. In fact $E=mc^2$ is a great example of where the principle of naturalness works. Of course, naturalness cannot explain why $E=mc^2$ at any deep level, but it does serve as an intuitive guide towards the formula.
May
17
comment Why isn't it $E \approx 27.642 \times mc^2$?
@badroit - essentially you are right about $E=mc^2$ being surprising if you've chosen units of mass, length and time consistently. I've elaborated on how exactly this works in my answer below.
May
16
answered Why isn't it $E \approx 27.642 \times mc^2$?